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I am doing the telescoping technique with partial sum of this infinite series because I want to find an upper bound of its partial sum. \begin{equation} \sum_{n=1}^{\infty} \frac{1}{n(n+2)} \end{equation} First I set $ \frac{1}{n(n+2)}= \frac{A}{n}+\frac{B}{n+2}$ and solve to get $A=\frac{1}{2}$ and $B=-\frac{1}{2}$. So I'm finding $\sum_{n=1}^{\infty}(\frac{1}{2n}-\frac{1}{2n+4})$ and proceed with the partial sum expansion. At the end I am left with $S_N=\frac{1}{2}+\frac{1}{4}-\frac{1}{2N+4}$ by cancelling terms in between and then $\lim_{N \to \infty}S_N=\frac{3}{4}$.

Am I doing it correctly? Thank you for your time.

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    $\begingroup$ yep $\endgroup$ – tilper Jan 3 at 14:04
  • $\begingroup$ Thanks @tilper. $\endgroup$ – Allorja Jan 3 at 14:06
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    $\begingroup$ Check again on your partial sums. I think you missed another negative term (though it tends to zero). $\endgroup$ – JavaMan Jan 3 at 14:07
  • $\begingroup$ Yeah I know I am missing a negative term because it cancels with a positive one in the following second pair of parentheses but I can always cancel while tending to infinity and it gets reallyyyy small. Thank you though. $\endgroup$ – Allorja Jan 3 at 14:10
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$$\sum^{\infty}_{n=1}\frac{1}{n(n+2)} = \frac{1}{2}\sum^{\infty}_{n=1}\bigg[\frac{1}{n}-\frac{1}{n+2}\bigg]$$

$$ = \frac{1}{2}\sum^{\infty}_{n=1}\bigg[\bigg(\frac{1}{n}-\frac{1}{n+1}\bigg)+\bigg(\frac{1}{n+1}-\frac{1}{n+2}\bigg)\bigg]$$

$$ = \frac{1}{2}\bigg[\frac{1}{1}+\frac{1}{2}\bigg] = \frac{3}{4}$$

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    $\begingroup$ Sorry but did you read the question? (The same would go for four upvoters...) $\endgroup$ – Did Jan 3 at 15:22
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In case you're interested, you can also compute this series using a certain representation of the digamma function $\psi(z+1)=-\gamma+\sum_{n\ge 1} \frac{z}{n(n+z)}$. We are then looking for $\frac{\psi(3)+\gamma}{2}$. Using the following integral representation of the digamma function, $$\psi(s+1)=-\gamma+\int_0^1 \frac{1-x^s}{1-x}\,dx$$ we find that $\psi(3)=-\gamma+\frac{3}{2}$. Your sum is then $$\sum_{n\ge 1} \frac{1}{n(n+2)}=\frac{3}{4}$$

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  • $\begingroup$ Thank you for your answer and time but I have not reached this level of mathematics yet so I don't have a clue about this way. But thank you though. $\endgroup$ – Allorja Jan 3 at 19:23

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