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Find general term of $1+frac{2!}{3}+frac{3!}{11}+frac{4!}{43}+frac{5!}{171}+…$ [closed]

4

2

$begingroup$

Find general term of $1+frac{2!}{3}+frac{3!}{11}+frac{4!}{43}+frac{5!}{171}+....$

However it has been ask to check convergence but how can i do that before knowing the general term. I can't see any pattern,comment quickly!

sequences-and-series pattern-recognition

share|cite|improve this question

edited Mar 21 at 15:29

tatan

5,79962761

asked Mar 21 at 15:27

M DesmondM Desmond

3248

$endgroup$

closed as off-topic by Saad, mrtaurho, Cesareo, MickG, GNUSupporter 8964民主女神 地下教會 Mar 22 at 11:32

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, mrtaurho, Cesareo, MickG, GNUSupporter 8964民主女神 地下教會

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
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  The denominators differ by powers of $2$. The differences are: $2, 8, 32, 128$. Perhaps this helps point you in the right direction.
  $endgroup$
  – Hyperion
  Mar 21 at 15:30
  

  
  
  
  


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  $begingroup$
  For the denominator: $frac{2^{2n + 1} + 1}{3}$
  $endgroup$
  – cansomeonehelpmeout
  Mar 21 at 15:32
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How did you get that @cansomehelpmeout
  $endgroup$
  – NewBornMATH
  Mar 21 at 15:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @cansomeonehelpmeout $frac{2^{2n-1}+1}{3}$ (exponent $2nmathbf{-}1$), I should think. This way, for $n=1$, we have $frac{2^1+1}{3}=1$, for $n=2$ $frac{2^3+1}{3}=3$, for $n=3$ $frac{2^5+1}{3}=frac{32+1}{3}=11$, for $n=4$ $frac{2^7+1}{3}=frac{128+1}{3}=43$, and for $n=5$ $frac{2^9+1}{3}=frac{512+1}{3}=171$, which matches the sequence.
  $endgroup$
  – MickG
  Mar 22 at 11:08
  

  
  
  
  

add a comment | 

4

2

$begingroup$

Find general term of $1+frac{2!}{3}+frac{3!}{11}+frac{4!}{43}+frac{5!}{171}+....$

However it has been ask to check convergence but how can i do that before knowing the general term. I can't see any pattern,comment quickly!

sequences-and-series pattern-recognition

share|cite|improve this question

edited Mar 21 at 15:29

tatan

5,79962761

asked Mar 21 at 15:27

M DesmondM Desmond

3248

$endgroup$

closed as off-topic by Saad, mrtaurho, Cesareo, MickG, GNUSupporter 8964民主女神 地下教會 Mar 22 at 11:32

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, mrtaurho, Cesareo, MickG, GNUSupporter 8964民主女神 地下教會

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  The denominators differ by powers of $2$. The differences are: $2, 8, 32, 128$. Perhaps this helps point you in the right direction.
  $endgroup$
  – Hyperion
  Mar 21 at 15:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  For the denominator: $frac{2^{2n + 1} + 1}{3}$
  $endgroup$
  – cansomeonehelpmeout
  Mar 21 at 15:32
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How did you get that @cansomehelpmeout
  $endgroup$
  – NewBornMATH
  Mar 21 at 15:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @cansomeonehelpmeout $frac{2^{2n-1}+1}{3}$ (exponent $2nmathbf{-}1$), I should think. This way, for $n=1$, we have $frac{2^1+1}{3}=1$, for $n=2$ $frac{2^3+1}{3}=3$, for $n=3$ $frac{2^5+1}{3}=frac{32+1}{3}=11$, for $n=4$ $frac{2^7+1}{3}=frac{128+1}{3}=43$, and for $n=5$ $frac{2^9+1}{3}=frac{512+1}{3}=171$, which matches the sequence.
  $endgroup$
  – MickG
  Mar 22 at 11:08
  

  
  
  
  

add a comment | 

$begingroup$

Find general term of $1+frac{2!}{3}+frac{3!}{11}+frac{4!}{43}+frac{5!}{171}+....$

However it has been ask to check convergence but how can i do that before knowing the general term. I can't see any pattern,comment quickly!

sequences-and-series pattern-recognition

share|cite|improve this question

edited Mar 21 at 15:29

tatan

5,79962761

asked Mar 21 at 15:27

M DesmondM Desmond

3248

$endgroup$

share|cite|improve this question

edited Mar 21 at 15:29

tatan

5,79962761

asked Mar 21 at 15:27

M DesmondM Desmond

3248

share|cite|improve this question

edited Mar 21 at 15:29

tatan

5,79962761

asked Mar 21 at 15:27

M DesmondM Desmond

3248

edited Mar 21 at 15:29

tatan

5,79962761

edited Mar 21 at 15:29

tatan

5,79962761

asked Mar 21 at 15:27

M DesmondM Desmond

3248

asked Mar 21 at 15:27

M DesmondM Desmond

3248

2 Answers
2

active

oldest

votes

2

$begingroup$

Notice that the numerator comprises factorials increasing by $1$ in each successive term. For the denominator it requires a bit more observation. The difference between the denominators of successive terms is our cue to guess there's an exponential term involved.

First Guess: $2^{2n+1}+1$ because have a look at the differences, they are differences of $2, 8, 32, ldots$.

But clearly doing so gives us the denominators as $3, 9, 33,ldots$ which is thrice of what our actual denominators are so we divide by $3$ to get the desired general $n^{text{th}}$ term of the sequence.

$$S_n=sum_{k=0}^{n}dfrac{3(k+1)!}{2^{2k+1}+1}$$

share|cite|improve this answer

edited Mar 21 at 16:14

answered Mar 21 at 15:43

Paras KhoslaParas Khosla

3,208626

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The answer is correct but I discourage you to post just the answer without explanation (you may post is as a comment and not an answer) as it does not help the OP with his doubt ;-)
  $endgroup$
  – tatan
  Mar 21 at 15:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Anyways, you seem to be a genius at your age. All the best for the future!
  $endgroup$
  – tatan
  Mar 21 at 15:57
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @tatan I've added the explanation. Thanks a bunch for your kind words. Means a lot to me. and All the very best to you too. Cheers :))
  $endgroup$
  – Paras Khosla
  Mar 21 at 16:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @paras how old are you ?
  $endgroup$
  – M Desmond
  Mar 22 at 20:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Can you figure out a way to prove $2^{n+1}-1>2^n$ for n in $N$(natural numbers)
  $endgroup$
  – M Desmond
  Mar 22 at 20:20
  
  
  

  
  
  
  

 | 
show 1 more comment

2

$begingroup$

Hint

The numerator is easy.

For the denominator, see the succesive differences. If you still can't figure out see Arithmetico–geometric sequence.

share|cite|improve this answer

answered Mar 21 at 15:31

tatantatan

5,79962761

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Well i observe a recurrence relation $y_{n}=a y_{n-1}+b$ where $a$ and $b$ are constant and thus solving it i got the general term. Wikipedia says that arithmetico geometric series form the former type of recurrence but i am unable to see the agp series directly, can you show me that ?( I.e. to write down the series in terms of $ar,(a+d)r^2,(a+2d)r^3...$ and what are a,r d for this question ?)
  $endgroup$
  – M Desmond
  Mar 25 at 7:32
  
  
  

  
  
  
  

add a comment | 

2

$begingroup$

Notice that the numerator comprises factorials increasing by $1$ in each successive term. For the denominator it requires a bit more observation. The difference between the denominators of successive terms is our cue to guess there's an exponential term involved.

First Guess: $2^{2n+1}+1$ because have a look at the differences, they are differences of $2, 8, 32, ldots$.

But clearly doing so gives us the denominators as $3, 9, 33,ldots$ which is thrice of what our actual denominators are so we divide by $3$ to get the desired general $n^{text{th}}$ term of the sequence.

$$S_n=sum_{k=0}^{n}dfrac{3(k+1)!}{2^{2k+1}+1}$$

share|cite|improve this answer

edited Mar 21 at 16:14

answered Mar 21 at 15:43

Paras KhoslaParas Khosla

3,208626

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The answer is correct but I discourage you to post just the answer without explanation (you may post is as a comment and not an answer) as it does not help the OP with his doubt ;-)
  $endgroup$
  – tatan
  Mar 21 at 15:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Anyways, you seem to be a genius at your age. All the best for the future!
  $endgroup$
  – tatan
  Mar 21 at 15:57
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @tatan I've added the explanation. Thanks a bunch for your kind words. Means a lot to me. and All the very best to you too. Cheers :))
  $endgroup$
  – Paras Khosla
  Mar 21 at 16:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @paras how old are you ?
  $endgroup$
  – M Desmond
  Mar 22 at 20:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Can you figure out a way to prove $2^{n+1}-1>2^n$ for n in $N$(natural numbers)
  $endgroup$
  – M Desmond
  Mar 22 at 20:20
  
  
  

  
  
  
  

 | 
show 1 more comment

2

$begingroup$

Notice that the numerator comprises factorials increasing by $1$ in each successive term. For the denominator it requires a bit more observation. The difference between the denominators of successive terms is our cue to guess there's an exponential term involved.

First Guess: $2^{2n+1}+1$ because have a look at the differences, they are differences of $2, 8, 32, ldots$.

But clearly doing so gives us the denominators as $3, 9, 33,ldots$ which is thrice of what our actual denominators are so we divide by $3$ to get the desired general $n^{text{th}}$ term of the sequence.

$$S_n=sum_{k=0}^{n}dfrac{3(k+1)!}{2^{2k+1}+1}$$

share|cite|improve this answer

edited Mar 21 at 16:14

answered Mar 21 at 15:43

Paras KhoslaParas Khosla

3,208626

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The answer is correct but I discourage you to post just the answer without explanation (you may post is as a comment and not an answer) as it does not help the OP with his doubt ;-)
  $endgroup$
  – tatan
  Mar 21 at 15:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Anyways, you seem to be a genius at your age. All the best for the future!
  $endgroup$
  – tatan
  Mar 21 at 15:57
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @tatan I've added the explanation. Thanks a bunch for your kind words. Means a lot to me. and All the very best to you too. Cheers :))
  $endgroup$
  – Paras Khosla
  Mar 21 at 16:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @paras how old are you ?
  $endgroup$
  – M Desmond
  Mar 22 at 20:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Can you figure out a way to prove $2^{n+1}-1>2^n$ for n in $N$(natural numbers)
  $endgroup$
  – M Desmond
  Mar 22 at 20:20
  
  
  

  
  
  
  

 | 
show 1 more comment

$begingroup$

Notice that the numerator comprises factorials increasing by $1$ in each successive term. For the denominator it requires a bit more observation. The difference between the denominators of successive terms is our cue to guess there's an exponential term involved.

First Guess: $2^{2n+1}+1$ because have a look at the differences, they are differences of $2, 8, 32, ldots$.

But clearly doing so gives us the denominators as $3, 9, 33,ldots$ which is thrice of what our actual denominators are so we divide by $3$ to get the desired general $n^{text{th}}$ term of the sequence.

$$S_n=sum_{k=0}^{n}dfrac{3(k+1)!}{2^{2k+1}+1}$$

share|cite|improve this answer

edited Mar 21 at 16:14

answered Mar 21 at 15:43

Paras KhoslaParas Khosla

3,208626

$endgroup$

share|cite|improve this answer

edited Mar 21 at 16:14

answered Mar 21 at 15:43

Paras KhoslaParas Khosla

3,208626

answered Mar 21 at 15:43

Paras KhoslaParas Khosla

3,208626

answered Mar 21 at 15:43

Paras KhoslaParas Khosla

3,208626

2

$begingroup$

Hint

The numerator is easy.

For the denominator, see the succesive differences. If you still can't figure out see Arithmetico–geometric sequence.

share|cite|improve this answer

answered Mar 21 at 15:31

tatantatan

5,79962761

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Well i observe a recurrence relation $y_{n}=a y_{n-1}+b$ where $a$ and $b$ are constant and thus solving it i got the general term. Wikipedia says that arithmetico geometric series form the former type of recurrence but i am unable to see the agp series directly, can you show me that ?( I.e. to write down the series in terms of $ar,(a+d)r^2,(a+2d)r^3...$ and what are a,r d for this question ?)
  $endgroup$
  – M Desmond
  Mar 25 at 7:32
  
  
  

  
  
  
  

add a comment | 

2

$begingroup$

Hint

The numerator is easy.

For the denominator, see the succesive differences. If you still can't figure out see Arithmetico–geometric sequence.

share|cite|improve this answer

answered Mar 21 at 15:31

tatantatan

5,79962761

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Well i observe a recurrence relation $y_{n}=a y_{n-1}+b$ where $a$ and $b$ are constant and thus solving it i got the general term. Wikipedia says that arithmetico geometric series form the former type of recurrence but i am unable to see the agp series directly, can you show me that ?( I.e. to write down the series in terms of $ar,(a+d)r^2,(a+2d)r^3...$ and what are a,r d for this question ?)
  $endgroup$
  – M Desmond
  Mar 25 at 7:32
  
  
  

  
  
  
  

add a comment | 

$begingroup$

Hint

The numerator is easy.

For the denominator, see the succesive differences. If you still can't figure out see Arithmetico–geometric sequence.

share|cite|improve this answer

answered Mar 21 at 15:31

tatantatan

5,79962761

$endgroup$

share|cite|improve this answer

answered Mar 21 at 15:31

tatantatan

5,79962761

answered Mar 21 at 15:31

tatantatan

5,79962761

answered Mar 21 at 15:31

tatantatan

5,79962761