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If $sum a_n$ converges and $b_n=sumlimits_{k=n}^{infty}a_n $, prove that $sum frac{a_n}{b_n}$ diverges

What can be said about the convergence/divergence of $sum a_n/r_n$ when $sum a_n$ is convergent?Prove that the series $sum frac{a_n}{r_n}$ diverges.An important lemma involving diverging sequencesMembers divided by remainder series diverges$frac {a_{n+1}}{a_n} le frac {b_{n+1}}{b_n}$ If $sum_{n=1}^infty b_n$ converges then $sum_{n=1}^infty a_n$ converges as wellIf $ sum a_n$ diverges and $lambda_n to infty$, does the series $ sum lambda_na_n$ diverge?If $sum a_n b_n$ converges for all $(b_n)$ such that $b_n to 0$, then $sum |a_n|$ converges.If $sumlimits_{n=1}^infty a_n$ diverges, then $sumlimits_{n=1}^inftyexpleft(-sumlimits_{k=1}^n k a_kright)$ converges?If a ${a_n}$ diverges, so $a_n rightarrow + infty$, how to find sequence ${b_n}$ such that $sum |b_n|<infty$ but $sum |a_n||b_n|$ diverges?Show that if $sum b_n$ is a rearrangement of a series $sum a_n$ , and $a_n$ diverges to $infty$, then $sum b_n = infty$Convergence of $sumfrac{a_n}{b_n} $ and $ sum (frac{a_n}{b_n})^2 $ implies convergence of $ sumfrac{a_n}{a_n+b_n}$Convergence of infinite series for $sum {{a_n}} $ and $sum {{b_n}}$$a_n$ is convergent, $b_n$ bounded, prove $sum a_n b_n$ convergesProve that the series $sum frac{a_n}{r_n}$ diverges.

21

11

$begingroup$

Let $displaystyle sum a_n$ be convergent series of positive terms and set $displaystyle b_n=sum_{k=n}^{infty}a_n$ , then prove that $displaystylesum frac{a_n}{b_n}$ diverges.

I could see that ${b_n}$ is monotonically decreasing sequence converging to $0$ and I can write $displaystylesum frac{a_n}{b_n}=sumfrac{b_n-b_{n+1}}{b_n}$, how shall I proceed further?

real-analysis sequences-and-series analysis divergent-series

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edited Dec 24 '15 at 11:00

Yiorgos S. Smyrlis

63.6k1385165

asked Aug 11 '14 at 7:28

BhauryalBhauryal

3,2391237

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is this homework, though?
  $endgroup$
  – k.stm
  Aug 11 '14 at 7:33
  

  
  
  
  

add a comment | 

21

11

$begingroup$

Let $displaystyle sum a_n$ be convergent series of positive terms and set $displaystyle b_n=sum_{k=n}^{infty}a_n$ , then prove that $displaystylesum frac{a_n}{b_n}$ diverges.

I could see that ${b_n}$ is monotonically decreasing sequence converging to $0$ and I can write $displaystylesum frac{a_n}{b_n}=sumfrac{b_n-b_{n+1}}{b_n}$, how shall I proceed further?

real-analysis sequences-and-series analysis divergent-series

share|cite|improve this question

edited Dec 24 '15 at 11:00

Yiorgos S. Smyrlis

63.6k1385165

asked Aug 11 '14 at 7:28

BhauryalBhauryal

3,2391237

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is this homework, though?
  $endgroup$
  – k.stm
  Aug 11 '14 at 7:33
  

  
  
  
  

add a comment | 

$begingroup$

Let $displaystyle sum a_n$ be convergent series of positive terms and set $displaystyle b_n=sum_{k=n}^{infty}a_n$ , then prove that $displaystylesum frac{a_n}{b_n}$ diverges.

I could see that ${b_n}$ is monotonically decreasing sequence converging to $0$ and I can write $displaystylesum frac{a_n}{b_n}=sumfrac{b_n-b_{n+1}}{b_n}$, how shall I proceed further?

real-analysis sequences-and-series analysis divergent-series

share|cite|improve this question

edited Dec 24 '15 at 11:00

Yiorgos S. Smyrlis

63.6k1385165

asked Aug 11 '14 at 7:28

BhauryalBhauryal

3,2391237

$endgroup$

share|cite|improve this question

edited Dec 24 '15 at 11:00

Yiorgos S. Smyrlis

63.6k1385165

asked Aug 11 '14 at 7:28

BhauryalBhauryal

3,2391237

share|cite|improve this question

edited Dec 24 '15 at 11:00

Yiorgos S. Smyrlis

63.6k1385165

asked Aug 11 '14 at 7:28

BhauryalBhauryal

3,2391237

edited Dec 24 '15 at 11:00

Yiorgos S. Smyrlis

63.6k1385165

edited Dec 24 '15 at 11:00

Yiorgos S. Smyrlis

63.6k1385165

asked Aug 11 '14 at 7:28

BhauryalBhauryal

3,2391237

asked Aug 11 '14 at 7:28

BhauryalBhauryal

3,2391237

6 Answers
6

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10

$begingroup$

Sorry, my previous answer was not correct. A new tentative:

$$frac{b_{k+1}}{b_{k}}=1-frac{a_k}{b_k}$$
Hence
$$frac{b_{N+1}}{b_1}=prod_{k=1}^N{(1-frac{a_k}{b_k}})$$ and
$$log b_{N+1}-log b_1=sum_{k=1}^N log(1-frac{a_k}{b_k})$$
Now if the series $a_k/b_k$ is convergent, we have $a_k/b_k to 0$, and as $log(1-x)sim -x$ and the series have constant sign, this imply that the series $displaystyle log (1-frac{a_k}{b_k})$ is convergent, a contradiction as $log (b_{N+1}) to -infty$.

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edited Aug 11 '14 at 8:42

answered Aug 11 '14 at 7:44

KelennerKelenner

17.5k1830

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5

$begingroup$

Assuming $sum a_n=L$, for any $n$ big enough we must have:
$$a_n geq frac{1}{n}sum_{m>n}a_m,tag{1}$$
otherwise, assuming that $N$ is the greatest integer for which $(1)$ holds, we have:
$$L< a_N+frac{1}{N}(L-a_n)+frac{N-1}{(N+1)N}(L-a_n)+ldots = L,tag{2}$$
contradiction. This implies that for any $ngeq M$
$$left(1+frac{1}{n}right)a_ngeqfrac{1}{n}b_ntag{3}$$
holds, hence:
$$sum_{ngeq M}frac{a_n}{b_n}geqsum_{ngeq M}frac{1}{n+1},tag{4}$$
but the RHS of $(4)$ diverges.

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answered Aug 11 '14 at 8:03

Jack D'AurizioJack D'Aurizio

292k33284672

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5

$begingroup$

First note that ${b_n}_{ninmathbb N}$ is a decreasing sequence of positive numbers, which tends to zero.

We have that, for $n> m$
$$
frac{a_m}{b_m}+cdots+frac{a_n}{b_n}ge
frac{a_m}{b_m}+cdots+frac{a_n}{b_m}=frac{1}{b_m}(a_m+cdots+a_n)=frac{b_n-b_m}{b_m}=1-frac{b_n}{b_m}.
$$
Next, as $b_nsearrow 0$, choose $m_1,m_2,ldots,m_k,ldots$, so that
$$
frac{b_{m_{i+1}}}{b_{m_i}}<1/2.
$$
Then we have that
$$
sum_{n=1}^{m_k}frac{a_n}{b_n}gesum_{i=1}^{k-1}sum_{n=m_i+1}^{m_{i+1}}frac{a_n}{b_n}ge
sum_{i=1}^{k-1}left(1-frac{b_{m_i}}{b_{m_{i+1}}}right)gefrac{k-1}{2},
$$
and hence $displaystylesum_{n=1}^{infty}frac{a_n}{b_n}=infty$.

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edited Dec 24 '15 at 10:58

answered Aug 11 '14 at 7:44

Yiorgos S. SmyrlisYiorgos S. Smyrlis

63.6k1385165

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4

$begingroup$

For $m>n$ one has
$$begin{aligned}frac{a_n}{b_n}+cdotsfrac{a_m}{b_m}&ge frac{a_n}{b_n}+cdots +frac{a_m}{b_n}\
&=frac{b_n-b_{m+1}}{b_n} = 1-frac{b_{m+1}}{b_n}.
end{aligned}$$
Can you continue from here?

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answered Aug 11 '14 at 7:44

Quang HoangQuang Hoang

13.2k1233

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2

$begingroup$

For a sum $sum_{k=0}^{infty}c_k$ to converge its tail must converge to 0.

$$
lim_{n to infty} sum_{k=n}^{infty}c_k = 0
$$

i.e. for every $epsilon$ there exist a $n_0$ such that for all $n > n_0$

$$
left| sum_{k=n}^{infty}c_k right| < epsilon
$$

We can prove that there exists a $epsilon$ such that for all $n$ the sum is bigger than $epsilon$ and thus the sum diverges.

$$
sum_{k=n}^{infty}frac{a_k}{b_k} geq frac{1}{b_n}sum_{k=n}^{infty}a_k = 1 = epsilon
$$

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answered Aug 11 '14 at 8:50

vladimirmvladimirm

630514

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2

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First answer was wrong. Here is my new try :

If $underset{nto+infty}{lim} frac{a_n}{b_n}$ is not $ 0 $ or does not exist, $sum frac{a_n}{b_n}$ is directly divergent.

Otherwise we have $underset{nto+infty}{lim} frac{a_n}{b_n} = 0 $ so $b_{n+1} sim b_n $ . It implies $sum frac{a_n}{b_n} = sum frac{b_n-b_{n+1}}{b_n}$ and $sum frac{b_n-b_{n+1}}{b_{n+1}}$ have the same behavior thanks to limit comparison test.

Now, since $b_n$ is decreasing and tends to $0$ as $nto+infty$, we can use an integral comparison :

$$ sum_{n=1}^N frac{b_n-b_{n+1}}{b_{n+1}} ge sum_{n=1}^N int_{b_{n+1}}^{b_n} frac{dx}{x} = int_{b_{N+1}}^{b_1} frac{dx}{x} = ln left(frac{b_1}{b_{N+1}}right) underset{Nto+infty}{longrightarrow}+infty $$

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edited Aug 13 '14 at 12:17

answered Aug 12 '14 at 11:42

yultanyultan

1,25289

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  please explain $displaystyle frac{b_n-b_{n+1}}{b_n} ge int_{b_{n+1}}^{b_n} frac{dx}{x}$
  $endgroup$
  – Bhauryal
  Aug 12 '14 at 16:58
  

  
  
  
  


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  For any function $f$ positive on $[a,b], a < b $, we have (just make a quick drawing) : $$(b-a) cdot underset{xin [a,b]}{min} f(x) le int_a^b f(t) dt le (b-a)cdot underset{xin [a,b]}{max} f(x) $$ Another way to see it is that the mean value of the function $f$ over $[a,b]$ is between the minimum and the maximum value of $f$ on this interval.
  $endgroup$
  – yultan
  Aug 12 '14 at 19:48
  
  
  

  
  
  
  


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  Yes, but $x leq b_n$ and thus $frac{1}{x} geq frac{1}{b_n}$, but you need $frac{1}{x} leq frac{1}{b_n}$.
  $endgroup$
  – PhoemueX
  Aug 13 '14 at 8:14
  

  
  
  
  


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  $begingroup$
  Sorry I made a mistake. You are right it should be with $b_{n+1}$. Initial answer corrected, though.
  $endgroup$
  – yultan
  Aug 13 '14 at 11:56
  

  
  
  
  

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10

$begingroup$

Sorry, my previous answer was not correct. A new tentative:

$$frac{b_{k+1}}{b_{k}}=1-frac{a_k}{b_k}$$
Hence
$$frac{b_{N+1}}{b_1}=prod_{k=1}^N{(1-frac{a_k}{b_k}})$$ and
$$log b_{N+1}-log b_1=sum_{k=1}^N log(1-frac{a_k}{b_k})$$
Now if the series $a_k/b_k$ is convergent, we have $a_k/b_k to 0$, and as $log(1-x)sim -x$ and the series have constant sign, this imply that the series $displaystyle log (1-frac{a_k}{b_k})$ is convergent, a contradiction as $log (b_{N+1}) to -infty$.

share|cite|improve this answer

edited Aug 11 '14 at 8:42

answered Aug 11 '14 at 7:44

KelennerKelenner

17.5k1830

$endgroup$

add a comment | 

10

$begingroup$

Sorry, my previous answer was not correct. A new tentative:

$$frac{b_{k+1}}{b_{k}}=1-frac{a_k}{b_k}$$
Hence
$$frac{b_{N+1}}{b_1}=prod_{k=1}^N{(1-frac{a_k}{b_k}})$$ and
$$log b_{N+1}-log b_1=sum_{k=1}^N log(1-frac{a_k}{b_k})$$
Now if the series $a_k/b_k$ is convergent, we have $a_k/b_k to 0$, and as $log(1-x)sim -x$ and the series have constant sign, this imply that the series $displaystyle log (1-frac{a_k}{b_k})$ is convergent, a contradiction as $log (b_{N+1}) to -infty$.

share|cite|improve this answer

edited Aug 11 '14 at 8:42

answered Aug 11 '14 at 7:44

KelennerKelenner

17.5k1830

$endgroup$

add a comment | 

$begingroup$

Sorry, my previous answer was not correct. A new tentative:

$$frac{b_{k+1}}{b_{k}}=1-frac{a_k}{b_k}$$
Hence
$$frac{b_{N+1}}{b_1}=prod_{k=1}^N{(1-frac{a_k}{b_k}})$$ and
$$log b_{N+1}-log b_1=sum_{k=1}^N log(1-frac{a_k}{b_k})$$
Now if the series $a_k/b_k$ is convergent, we have $a_k/b_k to 0$, and as $log(1-x)sim -x$ and the series have constant sign, this imply that the series $displaystyle log (1-frac{a_k}{b_k})$ is convergent, a contradiction as $log (b_{N+1}) to -infty$.

share|cite|improve this answer

edited Aug 11 '14 at 8:42

answered Aug 11 '14 at 7:44

KelennerKelenner

17.5k1830

$endgroup$

share|cite|improve this answer

edited Aug 11 '14 at 8:42

answered Aug 11 '14 at 7:44

KelennerKelenner

17.5k1830

answered Aug 11 '14 at 7:44

KelennerKelenner

17.5k1830

answered Aug 11 '14 at 7:44

KelennerKelenner

17.5k1830

5

$begingroup$

Assuming $sum a_n=L$, for any $n$ big enough we must have:
$$a_n geq frac{1}{n}sum_{m>n}a_m,tag{1}$$
otherwise, assuming that $N$ is the greatest integer for which $(1)$ holds, we have:
$$L< a_N+frac{1}{N}(L-a_n)+frac{N-1}{(N+1)N}(L-a_n)+ldots = L,tag{2}$$
contradiction. This implies that for any $ngeq M$
$$left(1+frac{1}{n}right)a_ngeqfrac{1}{n}b_ntag{3}$$
holds, hence:
$$sum_{ngeq M}frac{a_n}{b_n}geqsum_{ngeq M}frac{1}{n+1},tag{4}$$
but the RHS of $(4)$ diverges.

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answered Aug 11 '14 at 8:03

Jack D'AurizioJack D'Aurizio

292k33284672

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add a comment | 

5

$begingroup$

Assuming $sum a_n=L$, for any $n$ big enough we must have:
$$a_n geq frac{1}{n}sum_{m>n}a_m,tag{1}$$
otherwise, assuming that $N$ is the greatest integer for which $(1)$ holds, we have:
$$L< a_N+frac{1}{N}(L-a_n)+frac{N-1}{(N+1)N}(L-a_n)+ldots = L,tag{2}$$
contradiction. This implies that for any $ngeq M$
$$left(1+frac{1}{n}right)a_ngeqfrac{1}{n}b_ntag{3}$$
holds, hence:
$$sum_{ngeq M}frac{a_n}{b_n}geqsum_{ngeq M}frac{1}{n+1},tag{4}$$
but the RHS of $(4)$ diverges.

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answered Aug 11 '14 at 8:03

Jack D'AurizioJack D'Aurizio

292k33284672

$endgroup$

add a comment | 

$begingroup$

Assuming $sum a_n=L$, for any $n$ big enough we must have:
$$a_n geq frac{1}{n}sum_{m>n}a_m,tag{1}$$
otherwise, assuming that $N$ is the greatest integer for which $(1)$ holds, we have:
$$L< a_N+frac{1}{N}(L-a_n)+frac{N-1}{(N+1)N}(L-a_n)+ldots = L,tag{2}$$
contradiction. This implies that for any $ngeq M$
$$left(1+frac{1}{n}right)a_ngeqfrac{1}{n}b_ntag{3}$$
holds, hence:
$$sum_{ngeq M}frac{a_n}{b_n}geqsum_{ngeq M}frac{1}{n+1},tag{4}$$
but the RHS of $(4)$ diverges.

share|cite|improve this answer

answered Aug 11 '14 at 8:03

Jack D'AurizioJack D'Aurizio

292k33284672

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share|cite|improve this answer

answered Aug 11 '14 at 8:03

Jack D'AurizioJack D'Aurizio

292k33284672

answered Aug 11 '14 at 8:03

Jack D'AurizioJack D'Aurizio

292k33284672

answered Aug 11 '14 at 8:03

Jack D'AurizioJack D'Aurizio

292k33284672

5

$begingroup$

First note that ${b_n}_{ninmathbb N}$ is a decreasing sequence of positive numbers, which tends to zero.

We have that, for $n> m$
$$
frac{a_m}{b_m}+cdots+frac{a_n}{b_n}ge
frac{a_m}{b_m}+cdots+frac{a_n}{b_m}=frac{1}{b_m}(a_m+cdots+a_n)=frac{b_n-b_m}{b_m}=1-frac{b_n}{b_m}.
$$
Next, as $b_nsearrow 0$, choose $m_1,m_2,ldots,m_k,ldots$, so that
$$
frac{b_{m_{i+1}}}{b_{m_i}}<1/2.
$$
Then we have that
$$
sum_{n=1}^{m_k}frac{a_n}{b_n}gesum_{i=1}^{k-1}sum_{n=m_i+1}^{m_{i+1}}frac{a_n}{b_n}ge
sum_{i=1}^{k-1}left(1-frac{b_{m_i}}{b_{m_{i+1}}}right)gefrac{k-1}{2},
$$
and hence $displaystylesum_{n=1}^{infty}frac{a_n}{b_n}=infty$.

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edited Dec 24 '15 at 10:58

answered Aug 11 '14 at 7:44

Yiorgos S. SmyrlisYiorgos S. Smyrlis

63.6k1385165

$endgroup$

add a comment | 

5

$begingroup$

First note that ${b_n}_{ninmathbb N}$ is a decreasing sequence of positive numbers, which tends to zero.

We have that, for $n> m$
$$
frac{a_m}{b_m}+cdots+frac{a_n}{b_n}ge
frac{a_m}{b_m}+cdots+frac{a_n}{b_m}=frac{1}{b_m}(a_m+cdots+a_n)=frac{b_n-b_m}{b_m}=1-frac{b_n}{b_m}.
$$
Next, as $b_nsearrow 0$, choose $m_1,m_2,ldots,m_k,ldots$, so that
$$
frac{b_{m_{i+1}}}{b_{m_i}}<1/2.
$$
Then we have that
$$
sum_{n=1}^{m_k}frac{a_n}{b_n}gesum_{i=1}^{k-1}sum_{n=m_i+1}^{m_{i+1}}frac{a_n}{b_n}ge
sum_{i=1}^{k-1}left(1-frac{b_{m_i}}{b_{m_{i+1}}}right)gefrac{k-1}{2},
$$
and hence $displaystylesum_{n=1}^{infty}frac{a_n}{b_n}=infty$.

share|cite|improve this answer

edited Dec 24 '15 at 10:58

answered Aug 11 '14 at 7:44

Yiorgos S. SmyrlisYiorgos S. Smyrlis

63.6k1385165

$endgroup$

add a comment | 

$begingroup$

First note that ${b_n}_{ninmathbb N}$ is a decreasing sequence of positive numbers, which tends to zero.

We have that, for $n> m$
$$
frac{a_m}{b_m}+cdots+frac{a_n}{b_n}ge
frac{a_m}{b_m}+cdots+frac{a_n}{b_m}=frac{1}{b_m}(a_m+cdots+a_n)=frac{b_n-b_m}{b_m}=1-frac{b_n}{b_m}.
$$
Next, as $b_nsearrow 0$, choose $m_1,m_2,ldots,m_k,ldots$, so that
$$
frac{b_{m_{i+1}}}{b_{m_i}}<1/2.
$$
Then we have that
$$
sum_{n=1}^{m_k}frac{a_n}{b_n}gesum_{i=1}^{k-1}sum_{n=m_i+1}^{m_{i+1}}frac{a_n}{b_n}ge
sum_{i=1}^{k-1}left(1-frac{b_{m_i}}{b_{m_{i+1}}}right)gefrac{k-1}{2},
$$
and hence $displaystylesum_{n=1}^{infty}frac{a_n}{b_n}=infty$.

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edited Dec 24 '15 at 10:58

answered Aug 11 '14 at 7:44

Yiorgos S. SmyrlisYiorgos S. Smyrlis

63.6k1385165

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edited Dec 24 '15 at 10:58

answered Aug 11 '14 at 7:44

Yiorgos S. SmyrlisYiorgos S. Smyrlis

63.6k1385165

answered Aug 11 '14 at 7:44

Yiorgos S. SmyrlisYiorgos S. Smyrlis

63.6k1385165

answered Aug 11 '14 at 7:44

Yiorgos S. SmyrlisYiorgos S. Smyrlis

63.6k1385165

4

$begingroup$

For $m>n$ one has
$$begin{aligned}frac{a_n}{b_n}+cdotsfrac{a_m}{b_m}&ge frac{a_n}{b_n}+cdots +frac{a_m}{b_n}\
&=frac{b_n-b_{m+1}}{b_n} = 1-frac{b_{m+1}}{b_n}.
end{aligned}$$
Can you continue from here?

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answered Aug 11 '14 at 7:44

Quang HoangQuang Hoang

13.2k1233

$endgroup$

add a comment | 

4

$begingroup$

For $m>n$ one has
$$begin{aligned}frac{a_n}{b_n}+cdotsfrac{a_m}{b_m}&ge frac{a_n}{b_n}+cdots +frac{a_m}{b_n}\
&=frac{b_n-b_{m+1}}{b_n} = 1-frac{b_{m+1}}{b_n}.
end{aligned}$$
Can you continue from here?

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answered Aug 11 '14 at 7:44

Quang HoangQuang Hoang

13.2k1233

$endgroup$

add a comment | 

$begingroup$

For $m>n$ one has
$$begin{aligned}frac{a_n}{b_n}+cdotsfrac{a_m}{b_m}&ge frac{a_n}{b_n}+cdots +frac{a_m}{b_n}\
&=frac{b_n-b_{m+1}}{b_n} = 1-frac{b_{m+1}}{b_n}.
end{aligned}$$
Can you continue from here?

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answered Aug 11 '14 at 7:44

Quang HoangQuang Hoang

13.2k1233

$endgroup$

share|cite|improve this answer

answered Aug 11 '14 at 7:44

Quang HoangQuang Hoang

13.2k1233

answered Aug 11 '14 at 7:44

Quang HoangQuang Hoang

13.2k1233

answered Aug 11 '14 at 7:44

Quang HoangQuang Hoang

13.2k1233

2

$begingroup$

For a sum $sum_{k=0}^{infty}c_k$ to converge its tail must converge to 0.

$$
lim_{n to infty} sum_{k=n}^{infty}c_k = 0
$$

i.e. for every $epsilon$ there exist a $n_0$ such that for all $n > n_0$

$$
left| sum_{k=n}^{infty}c_k right| < epsilon
$$

We can prove that there exists a $epsilon$ such that for all $n$ the sum is bigger than $epsilon$ and thus the sum diverges.

$$
sum_{k=n}^{infty}frac{a_k}{b_k} geq frac{1}{b_n}sum_{k=n}^{infty}a_k = 1 = epsilon
$$

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answered Aug 11 '14 at 8:50

vladimirmvladimirm

630514

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add a comment | 

2

$begingroup$

For a sum $sum_{k=0}^{infty}c_k$ to converge its tail must converge to 0.

$$
lim_{n to infty} sum_{k=n}^{infty}c_k = 0
$$

i.e. for every $epsilon$ there exist a $n_0$ such that for all $n > n_0$

$$
left| sum_{k=n}^{infty}c_k right| < epsilon
$$

We can prove that there exists a $epsilon$ such that for all $n$ the sum is bigger than $epsilon$ and thus the sum diverges.

$$
sum_{k=n}^{infty}frac{a_k}{b_k} geq frac{1}{b_n}sum_{k=n}^{infty}a_k = 1 = epsilon
$$

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answered Aug 11 '14 at 8:50

vladimirmvladimirm

630514

$endgroup$

add a comment | 

$begingroup$

For a sum $sum_{k=0}^{infty}c_k$ to converge its tail must converge to 0.

$$
lim_{n to infty} sum_{k=n}^{infty}c_k = 0
$$

i.e. for every $epsilon$ there exist a $n_0$ such that for all $n > n_0$

$$
left| sum_{k=n}^{infty}c_k right| < epsilon
$$

We can prove that there exists a $epsilon$ such that for all $n$ the sum is bigger than $epsilon$ and thus the sum diverges.

$$
sum_{k=n}^{infty}frac{a_k}{b_k} geq frac{1}{b_n}sum_{k=n}^{infty}a_k = 1 = epsilon
$$

share|cite|improve this answer

answered Aug 11 '14 at 8:50

vladimirmvladimirm

630514

$endgroup$

share|cite|improve this answer

answered Aug 11 '14 at 8:50

vladimirmvladimirm

630514

answered Aug 11 '14 at 8:50

vladimirmvladimirm

630514

answered Aug 11 '14 at 8:50

vladimirmvladimirm

630514

2

$begingroup$

First answer was wrong. Here is my new try :

If $underset{nto+infty}{lim} frac{a_n}{b_n}$ is not $ 0 $ or does not exist, $sum frac{a_n}{b_n}$ is directly divergent.

Otherwise we have $underset{nto+infty}{lim} frac{a_n}{b_n} = 0 $ so $b_{n+1} sim b_n $ . It implies $sum frac{a_n}{b_n} = sum frac{b_n-b_{n+1}}{b_n}$ and $sum frac{b_n-b_{n+1}}{b_{n+1}}$ have the same behavior thanks to limit comparison test.

Now, since $b_n$ is decreasing and tends to $0$ as $nto+infty$, we can use an integral comparison :

$$ sum_{n=1}^N frac{b_n-b_{n+1}}{b_{n+1}} ge sum_{n=1}^N int_{b_{n+1}}^{b_n} frac{dx}{x} = int_{b_{N+1}}^{b_1} frac{dx}{x} = ln left(frac{b_1}{b_{N+1}}right) underset{Nto+infty}{longrightarrow}+infty $$

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edited Aug 13 '14 at 12:17

answered Aug 12 '14 at 11:42

yultanyultan

1,25289

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  please explain $displaystyle frac{b_n-b_{n+1}}{b_n} ge int_{b_{n+1}}^{b_n} frac{dx}{x}$
  $endgroup$
  – Bhauryal
  Aug 12 '14 at 16:58
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  For any function $f$ positive on $[a,b], a < b $, we have (just make a quick drawing) : $$(b-a) cdot underset{xin [a,b]}{min} f(x) le int_a^b f(t) dt le (b-a)cdot underset{xin [a,b]}{max} f(x) $$ Another way to see it is that the mean value of the function $f$ over $[a,b]$ is between the minimum and the maximum value of $f$ on this interval.
  $endgroup$
  – yultan
  Aug 12 '14 at 19:48
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, but $x leq b_n$ and thus $frac{1}{x} geq frac{1}{b_n}$, but you need $frac{1}{x} leq frac{1}{b_n}$.
  $endgroup$
  – PhoemueX
  Aug 13 '14 at 8:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry I made a mistake. You are right it should be with $b_{n+1}$. Initial answer corrected, though.
  $endgroup$
  – yultan
  Aug 13 '14 at 11:56
  

  
  
  
  

add a comment | 

2

$begingroup$

First answer was wrong. Here is my new try :

If $underset{nto+infty}{lim} frac{a_n}{b_n}$ is not $ 0 $ or does not exist, $sum frac{a_n}{b_n}$ is directly divergent.

Otherwise we have $underset{nto+infty}{lim} frac{a_n}{b_n} = 0 $ so $b_{n+1} sim b_n $ . It implies $sum frac{a_n}{b_n} = sum frac{b_n-b_{n+1}}{b_n}$ and $sum frac{b_n-b_{n+1}}{b_{n+1}}$ have the same behavior thanks to limit comparison test.

Now, since $b_n$ is decreasing and tends to $0$ as $nto+infty$, we can use an integral comparison :

$$ sum_{n=1}^N frac{b_n-b_{n+1}}{b_{n+1}} ge sum_{n=1}^N int_{b_{n+1}}^{b_n} frac{dx}{x} = int_{b_{N+1}}^{b_1} frac{dx}{x} = ln left(frac{b_1}{b_{N+1}}right) underset{Nto+infty}{longrightarrow}+infty $$

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edited Aug 13 '14 at 12:17

answered Aug 12 '14 at 11:42

yultanyultan

1,25289

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  please explain $displaystyle frac{b_n-b_{n+1}}{b_n} ge int_{b_{n+1}}^{b_n} frac{dx}{x}$
  $endgroup$
  – Bhauryal
  Aug 12 '14 at 16:58
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  For any function $f$ positive on $[a,b], a < b $, we have (just make a quick drawing) : $$(b-a) cdot underset{xin [a,b]}{min} f(x) le int_a^b f(t) dt le (b-a)cdot underset{xin [a,b]}{max} f(x) $$ Another way to see it is that the mean value of the function $f$ over $[a,b]$ is between the minimum and the maximum value of $f$ on this interval.
  $endgroup$
  – yultan
  Aug 12 '14 at 19:48
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, but $x leq b_n$ and thus $frac{1}{x} geq frac{1}{b_n}$, but you need $frac{1}{x} leq frac{1}{b_n}$.
  $endgroup$
  – PhoemueX
  Aug 13 '14 at 8:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry I made a mistake. You are right it should be with $b_{n+1}$. Initial answer corrected, though.
  $endgroup$
  – yultan
  Aug 13 '14 at 11:56
  

  
  
  
  

add a comment | 

$begingroup$

First answer was wrong. Here is my new try :

If $underset{nto+infty}{lim} frac{a_n}{b_n}$ is not $ 0 $ or does not exist, $sum frac{a_n}{b_n}$ is directly divergent.

Otherwise we have $underset{nto+infty}{lim} frac{a_n}{b_n} = 0 $ so $b_{n+1} sim b_n $ . It implies $sum frac{a_n}{b_n} = sum frac{b_n-b_{n+1}}{b_n}$ and $sum frac{b_n-b_{n+1}}{b_{n+1}}$ have the same behavior thanks to limit comparison test.

Now, since $b_n$ is decreasing and tends to $0$ as $nto+infty$, we can use an integral comparison :

$$ sum_{n=1}^N frac{b_n-b_{n+1}}{b_{n+1}} ge sum_{n=1}^N int_{b_{n+1}}^{b_n} frac{dx}{x} = int_{b_{N+1}}^{b_1} frac{dx}{x} = ln left(frac{b_1}{b_{N+1}}right) underset{Nto+infty}{longrightarrow}+infty $$

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edited Aug 13 '14 at 12:17

answered Aug 12 '14 at 11:42

yultanyultan

1,25289

$endgroup$

share|cite|improve this answer

edited Aug 13 '14 at 12:17

answered Aug 12 '14 at 11:42

yultanyultan

1,25289

answered Aug 12 '14 at 11:42

yultanyultan

1,25289

answered Aug 12 '14 at 11:42

yultanyultan

1,25289