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Proof of convergence for non-absolute sequence

Proving the convergence of $a_n = frac{n}{n+sqrt n}$Convergence of Monotone Sequence in Affine-Extended RealsNon-increasing Monotone Sequence Convergence ProofEvaluating $lim_{n to infty} sumlimits_{k=1}^n frac{1}{k 2^k} $Suppose that the function $f:Irightarrowmathbb{R}$ is monotonically increasing and bounded. Prove that the $lim_{xrightarrow a}f(x)$ existsCheck my Work: Series ConvergenceProving that the limit of the product of these two sequences is zeroA sufficient condition for a sequence to converge if arithmetic mean of the sequence converges?Convergence of sequence of real functionIf a sequence of absolute values is bounded, does it then converge?

0

$begingroup$

I've established that the bounded sequence $A_{n}=sumlimits_{k=1}^{n}|x_{k}|$ converges, since it is bounded and monotonically increasing.
Now I'm trying to prove that the sequence $B_{n}=sumlimits_{k=1}^{n}x_{k}$ (from the same $x_{k}$) is also convergent. I'm trying to use the following fact: $|B_{n}-B_{m}|=A_{n}-A_{m}$ for $n>m$. Now since $A_{n}$ is convergent I know there exists a $N in mathbb{N}$ such that when $n,m > mathbb{N}$ and $epsilon>0$ we have $|A_{n}-L|<epsilon$ and $|A_{m}-L|<epsilon$.

Can I then just say $|B_{n}-B_{m}|=A_{n}-A_{m} = A_{n}-L+L-A_{m} leq |A_{n}-L|+|A_{m}-L| leq 2epsilon$?

I'm a bit confused since I'm not quite sure if it is actually true that $A_{n}$ converges to $L$ in stead of $|x_{k}|$. Because I think that if $A_{n}$ converges to $L$, then I would suppose $x_{k}$ would have to tend to $0$, but I don't see why this would be the case.

As you are probably able to see, I'm confused and any suggestions would be more than welcome.

real-analysis sequences-and-series limits analysis

share|cite|improve this question

asked Mar 10 at 22:10

MathbeginnerMathbeginner

1738

$endgroup$

add a comment | 

0

$begingroup$

I've established that the bounded sequence $A_{n}=sumlimits_{k=1}^{n}|x_{k}|$ converges, since it is bounded and monotonically increasing.
Now I'm trying to prove that the sequence $B_{n}=sumlimits_{k=1}^{n}x_{k}$ (from the same $x_{k}$) is also convergent. I'm trying to use the following fact: $|B_{n}-B_{m}|=A_{n}-A_{m}$ for $n>m$. Now since $A_{n}$ is convergent I know there exists a $N in mathbb{N}$ such that when $n,m > mathbb{N}$ and $epsilon>0$ we have $|A_{n}-L|<epsilon$ and $|A_{m}-L|<epsilon$.

Can I then just say $|B_{n}-B_{m}|=A_{n}-A_{m} = A_{n}-L+L-A_{m} leq |A_{n}-L|+|A_{m}-L| leq 2epsilon$?

I'm a bit confused since I'm not quite sure if it is actually true that $A_{n}$ converges to $L$ in stead of $|x_{k}|$. Because I think that if $A_{n}$ converges to $L$, then I would suppose $x_{k}$ would have to tend to $0$, but I don't see why this would be the case.

As you are probably able to see, I'm confused and any suggestions would be more than welcome.

real-analysis sequences-and-series limits analysis

share|cite|improve this question

asked Mar 10 at 22:10

MathbeginnerMathbeginner

1738

$endgroup$

add a comment | 

$begingroup$

I've established that the bounded sequence $A_{n}=sumlimits_{k=1}^{n}|x_{k}|$ converges, since it is bounded and monotonically increasing.
Now I'm trying to prove that the sequence $B_{n}=sumlimits_{k=1}^{n}x_{k}$ (from the same $x_{k}$) is also convergent. I'm trying to use the following fact: $|B_{n}-B_{m}|=A_{n}-A_{m}$ for $n>m$. Now since $A_{n}$ is convergent I know there exists a $N in mathbb{N}$ such that when $n,m > mathbb{N}$ and $epsilon>0$ we have $|A_{n}-L|<epsilon$ and $|A_{m}-L|<epsilon$.

Can I then just say $|B_{n}-B_{m}|=A_{n}-A_{m} = A_{n}-L+L-A_{m} leq |A_{n}-L|+|A_{m}-L| leq 2epsilon$?

I'm a bit confused since I'm not quite sure if it is actually true that $A_{n}$ converges to $L$ in stead of $|x_{k}|$. Because I think that if $A_{n}$ converges to $L$, then I would suppose $x_{k}$ would have to tend to $0$, but I don't see why this would be the case.

As you are probably able to see, I'm confused and any suggestions would be more than welcome.

real-analysis sequences-and-series limits analysis

share|cite|improve this question

asked Mar 10 at 22:10

MathbeginnerMathbeginner

1738

$endgroup$

share|cite|improve this question

asked Mar 10 at 22:10

MathbeginnerMathbeginner

1738

share|cite|improve this question

asked Mar 10 at 22:10

MathbeginnerMathbeginner

1738

asked Mar 10 at 22:10

MathbeginnerMathbeginner

1738

asked Mar 10 at 22:10

MathbeginnerMathbeginner

1738

1 Answer
1

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1

$begingroup$

You should use Cauchy criterion.

For $varepsilon > 0$, because $(A_n)$ converges there exists $N$ such that for $p > q geq N$,
$$|A_p - A_q| leq varepsilon, text{ i.e. } sum_{k=q+1}^p |x_k| leq varepsilon$$

So for $p>q geq N$,
$$|B_p - B_q| = left| sum_{k=q+1}^p x_k right| leq sum_{k=q+1}^p |x_k| leq varepsilon$$

This proves that $(B_n)$ is a Cauchy sequence, and if your space is complete, it converges.

share|cite|improve this answer

edited Mar 10 at 22:27

answered Mar 10 at 22:15

TheSilverDoeTheSilverDoe

3,866112

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I see what you mean and how this would work. But does that also mean my approach is wrong?
  $endgroup$
  – Mathbeginner
  Mar 10 at 22:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You say that $|B_n - B_m|=A_n-A_m$, but this has no reason to be true in general...
  $endgroup$
  – TheSilverDoe
  Mar 10 at 22:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I would argue that $|B_{n}-B_{m}|=|sumlimits_{k=1}^{n} x_{k}-sumlimits_{k=1}^{m} x_{k}|=|sumlimits_{k=m+1}^{n} x_{k}|=sumlimits_{k=m+1}^{n} |x_{k}|=A_{n}-A_{m}$. Right?
  $endgroup$
  – Mathbeginner
  Mar 10 at 22:22
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $|sum x_k|$ is not the same thing as $sum |x_k|$. The basic fact about $|.|$ is the triangle inequality $|sum x_k| leq sum |x_k|$. If you never heard about that, it is normal you are confused with absolute convergence questions ;)
  $endgroup$
  – TheSilverDoe
  Mar 10 at 22:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You're right, I should have written $|B_{n}-B_{m}| leq A_{n}-A_{m}$, but then I could use the same road I did, right?
  $endgroup$
  – Mathbeginner
  Mar 10 at 22:26
  

  
  
  
  

 | 
show 2 more comments

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1

$begingroup$

You should use Cauchy criterion.

For $varepsilon > 0$, because $(A_n)$ converges there exists $N$ such that for $p > q geq N$,
$$|A_p - A_q| leq varepsilon, text{ i.e. } sum_{k=q+1}^p |x_k| leq varepsilon$$

So for $p>q geq N$,
$$|B_p - B_q| = left| sum_{k=q+1}^p x_k right| leq sum_{k=q+1}^p |x_k| leq varepsilon$$

This proves that $(B_n)$ is a Cauchy sequence, and if your space is complete, it converges.

share|cite|improve this answer

edited Mar 10 at 22:27

answered Mar 10 at 22:15

TheSilverDoeTheSilverDoe

3,866112

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I see what you mean and how this would work. But does that also mean my approach is wrong?
  $endgroup$
  – Mathbeginner
  Mar 10 at 22:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You say that $|B_n - B_m|=A_n-A_m$, but this has no reason to be true in general...
  $endgroup$
  – TheSilverDoe
  Mar 10 at 22:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I would argue that $|B_{n}-B_{m}|=|sumlimits_{k=1}^{n} x_{k}-sumlimits_{k=1}^{m} x_{k}|=|sumlimits_{k=m+1}^{n} x_{k}|=sumlimits_{k=m+1}^{n} |x_{k}|=A_{n}-A_{m}$. Right?
  $endgroup$
  – Mathbeginner
  Mar 10 at 22:22
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $|sum x_k|$ is not the same thing as $sum |x_k|$. The basic fact about $|.|$ is the triangle inequality $|sum x_k| leq sum |x_k|$. If you never heard about that, it is normal you are confused with absolute convergence questions ;)
  $endgroup$
  – TheSilverDoe
  Mar 10 at 22:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You're right, I should have written $|B_{n}-B_{m}| leq A_{n}-A_{m}$, but then I could use the same road I did, right?
  $endgroup$
  – Mathbeginner
  Mar 10 at 22:26
  

  
  
  
  

 | 
show 2 more comments

1

$begingroup$

You should use Cauchy criterion.

For $varepsilon > 0$, because $(A_n)$ converges there exists $N$ such that for $p > q geq N$,
$$|A_p - A_q| leq varepsilon, text{ i.e. } sum_{k=q+1}^p |x_k| leq varepsilon$$

So for $p>q geq N$,
$$|B_p - B_q| = left| sum_{k=q+1}^p x_k right| leq sum_{k=q+1}^p |x_k| leq varepsilon$$

This proves that $(B_n)$ is a Cauchy sequence, and if your space is complete, it converges.

share|cite|improve this answer

edited Mar 10 at 22:27

answered Mar 10 at 22:15

TheSilverDoeTheSilverDoe

3,866112

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I see what you mean and how this would work. But does that also mean my approach is wrong?
  $endgroup$
  – Mathbeginner
  Mar 10 at 22:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You say that $|B_n - B_m|=A_n-A_m$, but this has no reason to be true in general...
  $endgroup$
  – TheSilverDoe
  Mar 10 at 22:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I would argue that $|B_{n}-B_{m}|=|sumlimits_{k=1}^{n} x_{k}-sumlimits_{k=1}^{m} x_{k}|=|sumlimits_{k=m+1}^{n} x_{k}|=sumlimits_{k=m+1}^{n} |x_{k}|=A_{n}-A_{m}$. Right?
  $endgroup$
  – Mathbeginner
  Mar 10 at 22:22
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $|sum x_k|$ is not the same thing as $sum |x_k|$. The basic fact about $|.|$ is the triangle inequality $|sum x_k| leq sum |x_k|$. If you never heard about that, it is normal you are confused with absolute convergence questions ;)
  $endgroup$
  – TheSilverDoe
  Mar 10 at 22:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You're right, I should have written $|B_{n}-B_{m}| leq A_{n}-A_{m}$, but then I could use the same road I did, right?
  $endgroup$
  – Mathbeginner
  Mar 10 at 22:26
  

  
  
  
  

 | 
show 2 more comments

$begingroup$

You should use Cauchy criterion.

For $varepsilon > 0$, because $(A_n)$ converges there exists $N$ such that for $p > q geq N$,
$$|A_p - A_q| leq varepsilon, text{ i.e. } sum_{k=q+1}^p |x_k| leq varepsilon$$

So for $p>q geq N$,
$$|B_p - B_q| = left| sum_{k=q+1}^p x_k right| leq sum_{k=q+1}^p |x_k| leq varepsilon$$

This proves that $(B_n)$ is a Cauchy sequence, and if your space is complete, it converges.

share|cite|improve this answer

edited Mar 10 at 22:27

answered Mar 10 at 22:15

TheSilverDoeTheSilverDoe

3,866112

$endgroup$

share|cite|improve this answer

edited Mar 10 at 22:27

answered Mar 10 at 22:15

TheSilverDoeTheSilverDoe

3,866112

answered Mar 10 at 22:15

TheSilverDoeTheSilverDoe

3,866112

answered Mar 10 at 22:15

TheSilverDoeTheSilverDoe

3,866112