Getting $f(x) =sum_{n=2}^{infty} dfrac{x^n}{n(n-1)}$, $|x|<1$. [duplicate]

1

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This question already has an answer here:

• 
  What is the sum of the power series: $sum_{k=2}^inftyfrac{x^k}{k(k-1)}$?
  
  2 answers
  
  

In its convergence range, we have that the function $f(x)$ has a Taylor series shown in this picture.

What is $f(x)$ itself?

Here is the Taylor series at $x=0$ for the function:

$$
f(x)=sum_{n=2}^{infty} dfrac{x^n}{n(n-1)}
$$

I'm not sure about the $1$ in $n-1$.

calculus taylor-expansion

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edited Jan 1 at 21:07

mrtaurho

4,05921234

asked Jan 1 at 19:28

mahsa.e.1378mahsa.e.1378

71

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marked as duplicate by Martin R, mrtaurho, Community♦ Jan 1 at 20:19

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Use partial fractions to simplify the sum.
  $endgroup$
  – D.B.
  Jan 1 at 19:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Linked images of problems are frowned upon here, IIRC. Check out the FAQ: math.meta.stackexchange.com/questions/107/…
  $endgroup$
  – LoveTooNap29
  Jan 1 at 19:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Another one: math.stackexchange.com/q/18532/42969 – Both found with Approach0
  $endgroup$
  – Martin R
  Jan 1 at 19:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  yes , thanks a lot
  $endgroup$
  – mahsa.e.1378
  Jan 1 at 20:09
  

  
  
  
  

add a comment | 

1

$begingroup$

This question already has an answer here:

• 
  What is the sum of the power series: $sum_{k=2}^inftyfrac{x^k}{k(k-1)}$?
  
  2 answers
  
  

In its convergence range, we have that the function $f(x)$ has a Taylor series shown in this picture.

What is $f(x)$ itself?

Here is the Taylor series at $x=0$ for the function:

$$
f(x)=sum_{n=2}^{infty} dfrac{x^n}{n(n-1)}
$$

I'm not sure about the $1$ in $n-1$.

calculus taylor-expansion

share|cite|improve this question

edited Jan 1 at 21:07

mrtaurho

4,05921234

asked Jan 1 at 19:28

mahsa.e.1378mahsa.e.1378

71

$endgroup$

marked as duplicate by Martin R, mrtaurho, Community♦ Jan 1 at 20:19

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Use partial fractions to simplify the sum.
  $endgroup$
  – D.B.
  Jan 1 at 19:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Linked images of problems are frowned upon here, IIRC. Check out the FAQ: math.meta.stackexchange.com/questions/107/…
  $endgroup$
  – LoveTooNap29
  Jan 1 at 19:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Another one: math.stackexchange.com/q/18532/42969 – Both found with Approach0
  $endgroup$
  – Martin R
  Jan 1 at 19:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  yes , thanks a lot
  $endgroup$
  – mahsa.e.1378
  Jan 1 at 20:09
  

  
  
  
  

add a comment | 

1

1

1

$begingroup$

This question already has an answer here:

• 
  What is the sum of the power series: $sum_{k=2}^inftyfrac{x^k}{k(k-1)}$?
  
  2 answers
  
  

In its convergence range, we have that the function $f(x)$ has a Taylor series shown in this picture.

What is $f(x)$ itself?

Here is the Taylor series at $x=0$ for the function:

$$
f(x)=sum_{n=2}^{infty} dfrac{x^n}{n(n-1)}
$$

I'm not sure about the $1$ in $n-1$.

calculus taylor-expansion

share|cite|improve this question

edited Jan 1 at 21:07

mrtaurho

4,05921234

asked Jan 1 at 19:28

mahsa.e.1378mahsa.e.1378

71

$endgroup$

This question already has an answer here:

• 
  What is the sum of the power series: $sum_{k=2}^inftyfrac{x^k}{k(k-1)}$?
  
  2 answers
  
  

In its convergence range, we have that the function $f(x)$ has a Taylor series shown in this picture.

What is $f(x)$ itself?

Here is the Taylor series at $x=0$ for the function:

$$
f(x)=sum_{n=2}^{infty} dfrac{x^n}{n(n-1)}
$$

I'm not sure about the $1$ in $n-1$.

This question already has an answer here:

• 
  What is the sum of the power series: $sum_{k=2}^inftyfrac{x^k}{k(k-1)}$?
  
  2 answers
  
  

calculus taylor-expansion

calculus taylor-expansion

share|cite|improve this question

edited Jan 1 at 21:07

mrtaurho

4,05921234

asked Jan 1 at 19:28

mahsa.e.1378mahsa.e.1378

71

share|cite|improve this question

edited Jan 1 at 21:07

mrtaurho

4,05921234

asked Jan 1 at 19:28

mahsa.e.1378mahsa.e.1378

71

share|cite|improve this question

share|cite|improve this question

edited Jan 1 at 21:07

mrtaurho

4,05921234

edited Jan 1 at 21:07

mrtaurho

4,05921234

edited Jan 1 at 21:07

mrtaurho

4,05921234

4,05921234

asked Jan 1 at 19:28

mahsa.e.1378mahsa.e.1378

71

asked Jan 1 at 19:28

mahsa.e.1378mahsa.e.1378

71

asked Jan 1 at 19:28

mahsa.e.1378mahsa.e.1378

71

71

marked as duplicate by Martin R, mrtaurho, Community♦ Jan 1 at 20:19

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

marked as duplicate by Martin R, mrtaurho, Community♦ Jan 1 at 20:19

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Use partial fractions to simplify the sum.
  $endgroup$
  – D.B.
  Jan 1 at 19:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Linked images of problems are frowned upon here, IIRC. Check out the FAQ: math.meta.stackexchange.com/questions/107/…
  $endgroup$
  – LoveTooNap29
  Jan 1 at 19:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Another one: math.stackexchange.com/q/18532/42969 – Both found with Approach0
  $endgroup$
  – Martin R
  Jan 1 at 19:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  yes , thanks a lot
  $endgroup$
  – mahsa.e.1378
  Jan 1 at 20:09
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Use partial fractions to simplify the sum.
  $endgroup$
  – D.B.
  Jan 1 at 19:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Linked images of problems are frowned upon here, IIRC. Check out the FAQ: math.meta.stackexchange.com/questions/107/…
  $endgroup$
  – LoveTooNap29
  Jan 1 at 19:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Another one: math.stackexchange.com/q/18532/42969 – Both found with Approach0
  $endgroup$
  – Martin R
  Jan 1 at 19:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  yes , thanks a lot
  $endgroup$
  – mahsa.e.1378
  Jan 1 at 20:09
  

  
  
  
  

1

1

$begingroup$
Use partial fractions to simplify the sum.
$endgroup$
– D.B.
Jan 1 at 19:31

$begingroup$
Use partial fractions to simplify the sum.
$endgroup$
– D.B.
Jan 1 at 19:31

$begingroup$
Linked images of problems are frowned upon here, IIRC. Check out the FAQ: math.meta.stackexchange.com/questions/107/…
$endgroup$
– LoveTooNap29
Jan 1 at 19:31

$begingroup$
Linked images of problems are frowned upon here, IIRC. Check out the FAQ: math.meta.stackexchange.com/questions/107/…
$endgroup$
– LoveTooNap29
Jan 1 at 19:31

$begingroup$
Another one: math.stackexchange.com/q/18532/42969 – Both found with Approach0
$endgroup$
– Martin R
Jan 1 at 19:47

$begingroup$
Another one: math.stackexchange.com/q/18532/42969 – Both found with Approach0
$endgroup$
– Martin R
Jan 1 at 19:47

$begingroup$
yes , thanks a lot
$endgroup$
– mahsa.e.1378
Jan 1 at 20:09

$begingroup$
yes , thanks a lot
$endgroup$
– mahsa.e.1378
Jan 1 at 20:09

add a comment | 

1 Answer
1

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Hint. One may recall that
$$
sum_{n=1}^inftyfrac{x^n}{n}=-lnleft(1-x right),quad |x|<1.
$$ Then writing
$$
frac{x^n}{n(n-1)}=xfrac{x^{n-1}}{n-1}-frac{x^n}{n},quad nge2,
$$ may help.

share|cite|improve this answer

answered Jan 1 at 19:33

Olivier OloaOlivier Oloa

108k17176293

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add a comment | 

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

3

$begingroup$

Hint. One may recall that
$$
sum_{n=1}^inftyfrac{x^n}{n}=-lnleft(1-x right),quad |x|<1.
$$ Then writing
$$
frac{x^n}{n(n-1)}=xfrac{x^{n-1}}{n-1}-frac{x^n}{n},quad nge2,
$$ may help.

share|cite|improve this answer

answered Jan 1 at 19:33

Olivier OloaOlivier Oloa

108k17176293

$endgroup$

add a comment | 

3

$begingroup$

Hint. One may recall that
$$
sum_{n=1}^inftyfrac{x^n}{n}=-lnleft(1-x right),quad |x|<1.
$$ Then writing
$$
frac{x^n}{n(n-1)}=xfrac{x^{n-1}}{n-1}-frac{x^n}{n},quad nge2,
$$ may help.

share|cite|improve this answer

answered Jan 1 at 19:33

Olivier OloaOlivier Oloa

108k17176293

$endgroup$

add a comment | 

3

3

3

$begingroup$

Hint. One may recall that
$$
sum_{n=1}^inftyfrac{x^n}{n}=-lnleft(1-x right),quad |x|<1.
$$ Then writing
$$
frac{x^n}{n(n-1)}=xfrac{x^{n-1}}{n-1}-frac{x^n}{n},quad nge2,
$$ may help.

share|cite|improve this answer

answered Jan 1 at 19:33

Olivier OloaOlivier Oloa

108k17176293

$endgroup$

Hint. One may recall that
$$
sum_{n=1}^inftyfrac{x^n}{n}=-lnleft(1-x right),quad |x|<1.
$$ Then writing
$$
frac{x^n}{n(n-1)}=xfrac{x^{n-1}}{n-1}-frac{x^n}{n},quad nge2,
$$ may help.

share|cite|improve this answer

answered Jan 1 at 19:33

Olivier OloaOlivier Oloa

108k17176293

share|cite|improve this answer

share|cite|improve this answer

answered Jan 1 at 19:33

Olivier OloaOlivier Oloa

108k17176293

answered Jan 1 at 19:33

Olivier OloaOlivier Oloa

108k17176293

answered Jan 1 at 19:33

Olivier OloaOlivier Oloa

108k17176293

108k17176293

add a comment | 

add a comment | 

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