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Let

$\lim_{n\rightarrow \infty }\sum_{k=2}^{n}\frac{1}{k^3-k}$

I converted the above series into following form by partial fraction method:

$\sum_{k=2}^{n}\frac{1}{k^3-k} =\sum_{k=2}^{n}\frac{-1}{k}+\frac{1}{2} \sum_{k=2}^{n}\frac{1}{(k+1)(k-1)}$

The sum of the second series can be done using the following method: $\frac{1}{4}\sum_{k=2}^{n}\frac{(k+1)-(k-1)}{(k+1)(k-1)}=\frac{1}{4}\sum_{k=2}^{n}[\frac{1}{k-1}-\frac{1}{k+1}]$

Which, then can be easily solved. But the first series clearly a divergent series. How can I then be able to solve the problem? The answer is given as one fourth.

Masacroso
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userNoOne
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2 Answers2

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$$\frac{1}{k^3-k}=\frac{1}{(k-1)k(k+1)}=\frac{1}{2}\left[\frac{1}{(k-1)k}-\frac{1}{k(k+1)}\right] $$ which is a telescopic term, hence $$ \sum_{k=2}^{n}\frac{1}{k^3-k}=\frac{1}{2}\left[\frac{1}{(2-1)2}-\frac{1}{n(n+1)}\right] $$ and the limit as $n\to +\infty$ is $\frac{1}{4}$.

Jack D'Aurizio
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  • Thank you so much sir. – userNoOne Jan 24 '18 at 11:05
  • No Matter how many problems I solve of this type, I get confused. If you can suggest any reference from where I can practice these problems where we have some series and we have to take the limit of the sum, it will be very beneficial for me. – userNoOne Jan 24 '18 at 11:07
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    @RAHUlJHa: the first section of my notes (https://drive.google.com/file/d/0BxKdOVsjsuEwdjBEM1dpRkhMa2s/view) is devoted to creative telescoping and similar problems. – Jack D'Aurizio Jan 24 '18 at 11:16
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    That's exactly what I was looking for. Thank you so much. Not only telescoping, now I am going study other topics also from your notes. They are just awesome. – userNoOne Jan 24 '18 at 11:25
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$$\lim_{n\rightarrow \infty }\sum_{k=2}^{n}\frac{1}{k^3-k}$$
$$ =\lim_{n\rightarrow \infty }\sum_{k=2}^{n}\frac{1}{k (k+1)(k-1)}$$
$$ =\lim_{n\rightarrow \infty }\sum_{k=2}^{n}\frac{(k+1)-(k-1)}{k (k+1)(k-1)}$$
$$ =\lim_{n\rightarrow \infty }\frac{1}{2}\left[\frac{1}{(k-1)k}-\frac{1}{k(k+1)}\right]$$
$$= \lim_{n\rightarrow \infty }\frac{1}{2}\left[\frac{1}{(2-1)2}-\frac{1}{n(n+1)}\right]= \frac{1}{4}$$

SDA OFFICIAL
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