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How to calculate the sum: $\sum_{n=1}^{\infty} \frac{1}{n(n+3)}$ ?
I know the sum converges because it is a positive sum for every $n$ and it is smaller than $\sum_{n=1}^{\infty} \frac{1}{n(n+1)}$ that converges and equals $1$. I need a direction...

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    $\begingroup$ Have you tried to use partial fractions to break $\frac{1}{n(n+3)}$ into a sum of the form $\frac{A}{n} + \frac{B}{n+3}$? $\endgroup$ – Trevor Norton Apr 13 '16 at 14:41
  • $\begingroup$ Very elementary and correct answers were given so I will not answer, but just so you know for future: There is a powerful method of summing series using complex integration. The so called "redisue theorem" is a poweful tool for integrating analytic functions on contrours in complex plane. But this can be reverted: if you can find a function such that its residues coincide with terms of your series, you can convert problem of summing the residues to problem of calculating some integral! This can be done in this case. $\endgroup$ – Blazej Apr 13 '16 at 14:53
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Using partial fraction decomposition $$ \frac{1}{n(n+3)} = \frac{1}{3} \cdot \frac{1}{n} - \frac{1}{3} \cdot \frac{1}{n+3} $$ convince yourself that for any $f$, we have $$ \sum_{n=1}^{m} \left(f(n)-f(n+3)\right) = f(1) + f(2) + f(3) - f(m+3) - f(m+2) - f(m+1) $$ and now take the limit of $m \to \infty$ getting $$ \sum_{n=1}^\infty \frac{1}{n(n+3)} = \frac{1}{3} \left(1 + \frac{1}{2} + \frac{1}{3} \right) = \frac{11}{18} $$

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$$\frac1{n(n+3)}=\frac13\left(\frac1n-\frac1{n+3}\right)\implies$$

$$\sum_{k=1}^n\frac1{k(k+3)}=\frac13\left(\frac11-\frac14+\frac12-\frac15+\frac13-\frac16+\frac14-\frac17+\ldots+\frac1n-\frac1{n+3}\right)=$$

$$\frac13\left(1+\frac12+\frac13-\frac1{n+1}-\frac1{n+2}-\frac1{n+3}\right)\xrightarrow[n\to\infty]{}\frac13\left(1+\frac56\right)=\frac{11}{18}$$

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  • $\begingroup$ @ThomasAndrews Thank you. I didn't understand what you wrote but your comment made me look again and I noticed the typos (-'s and +'s mixed there). Editing $\endgroup$ – DonAntonio Apr 13 '16 at 14:54
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\begin{align*} \frac{1}{n(n+3)} &= \frac{1}{3n}-\frac{1}{3(n+3)} \\ \sum_{n=1}^{\infty} \frac{1}{n(n+3)} &= \sum_{n=1}^{\infty} \left[ \frac{1}{3n}-\frac{1}{3(n+3)} \right] \\ &=\lim_{N\to \infty} \left[ \sum_{n=1}^{N} \frac{1}{3n}-\sum_{n=1}^{N} \frac{1}{3(n+3)} \right] \\ &=\sum_{n=1}^{3} \frac{1}{3n}- \lim_{N\to \infty} \sum_{n=1}^{3} \frac{1}{3(N+n)} \\ &=\frac{11}{18} \end{align*}

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  • $\begingroup$ This is a very nice way to reach the sum! +1 $\endgroup$ – DonAntonio Apr 13 '16 at 14:48
  • $\begingroup$ Sorry for the necropost - How do you justify the second to last step? $$\lim_{N\to \infty} \left[ \sum_{n=1}^{N} \frac{1}{3n}-\sum_{n=1}^{N} \frac{1}{3(n+3)} \right] \\ =\sum_{n=1}^{3} \frac{1}{3n}- \lim_{N\to \infty} \sum_{n=1}^{3} \frac{1}{3(N+n)} \\ $$ $\endgroup$ – Axion004 Nov 13 '19 at 15:08
  • $\begingroup$ Just deletion for repeated parts (in red): \begin{align} \sum_{n=1}^{N} \frac{1}{3n} &= \frac{1}{3\cdot 1}+\frac{1}{3\cdot 2}+\frac{1}{3\cdot 3}+ \color{red}{\frac{1}{3\cdot 4}+\ldots+\frac{1}{3N}} \\ \sum_{n=1}^{N} \frac{1}{3(n+3)} &= \color{red}{\frac{1}{3\cdot 4}+\ldots+\frac{1}{3N}}+ \frac{1}{3(N+1)}+\frac{1}{3(N+2)}+\frac{1}{3(N+3)} \end{align} $\endgroup$ – Ng Chung Tak Nov 13 '19 at 18:58
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The partial sum decomposition of the term of your serie is : $\frac{\frac{1}{3}}{n}-\frac{\frac{1}{3}}{n+3}$. You recognise telescoping series : $\sum_{n=1}^{\infty} \frac{1}{n(n+3)}=\sum_{n=1}^{\infty} \frac{\frac{1}{3}}{n}-\frac{\frac{1}{3}}{n+3}=\frac{\frac{1}{3}}{1}+\frac{\frac{1}{3}}{2}+\frac{\frac{1}{3}}{3}=\frac{11}{18}$.

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  • $\begingroup$ Your decompisition took you to the sum of two divergent series... $\endgroup$ – DonAntonio Apr 13 '16 at 14:47
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    $\begingroup$ It is not a problem, since i don't decompose the sum, I just use the decomposition to recognise a telescoping serie... $\endgroup$ – Jennifer Apr 13 '16 at 14:48
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You can use double integral to calculate. Let $f(x)=\sum_{n=1}^\infty\frac{1}{n(n+3)}x^n$. Then $f(1)=\sum_{n=1}^\infty\frac{1}{n(n+3)}$. Clearly $$ f'(x)=\sum_{n=1}^\infty\frac{1}{n+3}x^{n-1}$$ and $$ (x^4f'(x))'=\sum_{n=1}^\infty x^{n+2}=\frac{x^3}{1-x}.$$ Thus \begin{eqnarray} f(1)&=&\int_0^1\left(\int_0^x\frac{1}{x^4}\frac{t^3}{1-t}dt\right)dx\\ &=&\int_0^1\left(\int_t^1\frac{1}{x^4}\frac{t^3}{1-t}dx\right)dt\\ &=&-\frac13\int_0^1\left(1-\frac{1}{t^3}\right)\frac{t^3}{1-t}dt\\ &=&\frac13\int_0^1(t^2+t+1)dt\\ &=&\frac{11}{18}. \end{eqnarray}

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A lot of these answers involve taking the limit as the term number approaches $\infty$, but this is pretty tedious sometimes, so I thought an alternative might help with the evaluation of the sum.

$$\sum_{n=1}^\infty \frac{1}{n(n+3)}=\sum_{n=1}^\infty \bigg(\frac{A}{n}+\frac{B}{n+3}\bigg)=\sum_{n=1}^\infty \bigg(\frac{1/3}{n}-\frac{1/3}{n+3}\bigg)=\frac{1}{3}\sum_{n=1}^\infty \bigg(\frac{1}{n}-\frac{1}{n+3}\bigg)=\frac13\bigg(\sum_{n=1}^\infty \frac{1}{n}\bigg)-\frac13\bigg(\sum_{n=1}^\infty \frac{1}{n+3}\bigg)=\frac13\bigg(\sum_{n=1}^\infty \frac{1}{n}\bigg)-\frac13\bigg(\sum_{n=4}^\infty \frac{1}{n}\bigg)$$$$=\frac13\bigg(\sum_{n=1}^3\frac{1}{n}+\sum_{n=4}^\infty \frac{1}{n}\bigg)-\frac13\bigg(\sum_{n=4}^\infty \frac{1}{n}\bigg)=\frac13\sum_{n=1}^3\frac{1}{n}=\frac13\bigg(\frac11+\frac12+\frac13\bigg)=$$$$\frac13\bigg(\frac66+\frac36+\frac26\bigg)=\frac13\bigg(\frac{11}6\bigg)=\frac{11}{18} $$

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