4

I don't understand why I can't get the telescopic sum after the partial fraction decomposition:

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

I have $\sum\limits_{n=2}^{\infty }\frac{1}{n^3-n}=\phi(n+1)-\phi(0).$

The answer should be $1/4$.

David G. Stork
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mezzaluna
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  • Please provide more details, for instance, what are your thoughts about the question, what you have tried, where you are stuck. – Sahiba Arora Jul 26 '17 at 19:44
  • Rewrite the $\frac{-1}{n}=\frac{-1}{2n}+\frac{-1}{2n}$ in the middle ... it is telescopic ! – Donald Splutterwit Jul 26 '17 at 19:48
  • A telescope series is of the from $a_i - a_{i+d}$. Yours is of the form $a_{n} + b_{n-1} + b_{n+1}$. The $a_i$ don't seem to bear any relation to the $b_i$ terms and because you add the $b_i$ terms they compound rather than telescope. But if you do as Donald Splutterwit suggests you note $a_n = -2b_n$ so you actually have $(b_{n+1} - b_{n}) - (b_n - b_{n-1}$ which does telescope. – fleablood Jul 26 '17 at 20:12

3 Answers3

3

Hint: We have

$$\frac{1}{n(n+1)(n-1)}=\frac{1}{2}\left(\frac{1}{n(n-1)}-\frac{1}{n(n+1)}\right)$$

and the sum telescopes.

Brevan Ellefsen
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Oscar Lanzi
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If you had stopped at the ${-1\over n}+{1\over2(n-1)}+{1\over2(n+1)}$ and rewritten it as

$${1\over n^3-n}={1\over2}\left({1\over n-1}-{1\over n} \right)-{1\over2}\left({1\over n}-{1\over n+1} \right)$$

then the series does telescope. Note, however, you need the sum to start at $n=2$ rather than $n=0$, since $1/(n^3-n)$ is undefined for $n=0$ and $n=1$. (An edit to the OP corrected the starting point while I was posting this answer.) The only term that remains after the telescopic cancellations is

$${1\over2}\left({1\over2-1}-{1\over2}\right)={1\over4}$$

Barry Cipra
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$$\sum _{ n=2 }^{ \infty } \frac { 1 }{ n^{ 3 }-n } =\sum _{ n=2 }^{ \infty } \frac { 1 }{ \left( n-1 \right) n\left( n+1 \right) } =\frac { 1 }{ 2 } \sum _{ n=2 }^{ \infty } \left[ \frac { 1 }{ \left( n-1 \right) } +\frac { 1 }{ \left( n+1 \right) } -\frac { 2 }{ n } \right] =\\ =\frac { 1 }{ 2 } \sum _{ n=2 }^{ \infty } \left[ \frac { 1 }{ \left( n-1 \right) } -\frac { 1 }{ n } +\frac { 1 }{ \left( n+1 \right) } -\frac { 1 }{ n } \right] =\color{red}{\frac { 1 }{ 4 }} $$

haqnatural
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  • How can you separate $\sum _{n=2}^\infty\frac1n$ from the rest when this sum is infinite? The same comment applies to the other sums. The third expression in your answer evaluates to $\frac12(\infty+\infty)-\infty$. – Théophile Jul 26 '17 at 20:01
  • @Théophile:Thanks for replying;I don't understand the fact that if the sum is infinite so we cannot separate the terms – mezzaluna Jul 26 '17 at 20:16
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    @mezzaluna Consider, for example, $\sum _{n=1}^\infty1$, which is infinite. So an expression like $\sum _{n=1}^\infty1 - \sum _{n=1}^\infty1$ is not well defined, because it is $\infty-\infty$. On the other hand, $\sum _{n=1}^\infty(1-1) = \sum _{n=1}^\infty0 = 0$. So we have to be very careful when separating components of infinite sums. – Théophile Jul 27 '17 at 14:29