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I try to split the summand into differences, but that seems to be a futile way in our case right here, because the numerator is $k$, instead of a given number.

A closely-related series, say $\sum_{1}^{\infty}\frac{2}{(k+1)(k+2)(k+3)}$, can however be directly tackled by writing its partial sum as $$\sum_{1}^{n}\frac{1}{(k+1)(k+2)} - \frac{1}{(k+2)(k+3)}.$$

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  • $\begingroup$ Could $$\frac{k}{(k+1)(k+2)(k+3)}=\frac{1}{(k+1) (k+3)}-\frac{1}{(k+1) (k+2)}+\frac{1}{(k+3) (k+2)}$$ (in which you find the related series) could be of any help ? $\endgroup$ Mar 20, 2015 at 12:11
  • $\begingroup$ Thanks. Seems reasonable, I am trying :) $\endgroup$
    – Yes
    Mar 20, 2015 at 12:13

6 Answers 6

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$$\begin{align}\sum \limits_{k=1}^{\infty}\frac k{(k+1)(k+2)(k+3)}&= \sum \limits_{k=1}^{\infty}\dfrac{2}{k+2} - \dfrac{3}{2(k+3)} - \dfrac{1}{2(k+1)} \\~\\&=\frac{3}{2}\left(\sum \limits_{k=1}^{\infty}\dfrac{1}{k+2} -\dfrac{1}{k+3}\right) + \frac{1}{2}\left(\sum \limits_{k=1}^{\infty}\dfrac{1}{k+2}-\frac{1}{k+1}\right) \\~\\ &= \frac{3}{2}\left(\frac{1}{1+2}\right) + \frac{1}{2}\left(-\frac{1}{1+1}\right) \\~\\ &=\frac{1}{4}\end{align}$$

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  • $\begingroup$ That is the right track. $\endgroup$
    – Yes
    Mar 20, 2015 at 12:16
  • $\begingroup$ Haha it was a fun telescoping sum $$\sum\limits_{r=1}^{\infty}f(r) - f(r+1) = f(1)$$ provided $\lim\limits_{r\to\infty }f(r) = 0$ $\endgroup$
    – AgentS
    Mar 20, 2015 at 12:18
  • $\begingroup$ Nice. I should know better than to give subtle hints. :) $\endgroup$
    – PM 2Ring
    Mar 20, 2015 at 12:21
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Hint: You can split $$\frac{k}{(k+1)(k+2)(k+3)}$$ into $$\frac{a}{k+1} + \frac{b}{k+2} + \frac{c}{k+3}$$ where a, b & c are simple numbers. You can find those numbers by performing the addition and equating coefficients of powers of $k$, which will give you a set of 3 simultaneous equations in a, b & c.

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  • $\begingroup$ Thank you. Your hint is never subtle :). Just appeared a little too late. Instead, you provide a systematic method to solve a class of such problems. So it is valuable, I believe, and it is valuable especially for other viewers. $\endgroup$
    – Yes
    Mar 20, 2015 at 12:24
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We know that, $$\sum_{1}^{\infty}x^k=\dfrac {x}{1-x},|x| <1$$ Differentiating with respect to x, $$\sum_{1}^{\infty}kx^{k-1}=\dfrac {1}{(1-x)^2},|x| <1$$ Multiply by x,then integrate with respect to x, $$\sum_{1}^{\infty}\dfrac{kx^{k+1}}{k+1}=\dfrac {x}{(1-x)}+\log{(1-x)},|x| <1$$ Integrate with respect to x, $$\sum_{1}^{\infty}\dfrac{kx^{k+2}}{(k+1)(k+2)}=-2x-2\log{(1-x)}+x\log{(1-x)},|x| <1$$ Integrate with respect to x, $$\sum_{1}^{\infty}\dfrac{kx^{k+3}}{(k+1)(k+2)(k+3)}=\frac32x-\frac54x^2+\frac32\log{(1-x)}-2x\log{(1-x)}+\frac12x^2\log{(1-x)},|x| <1$$ Now taking Limit for $x$ tending to 1, We get, $$\sum_{1}^{\infty}\dfrac{k}{(k+1)(k+2)(k+3)}=\frac14$$

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  • $\begingroup$ I like such approaches! nice (+1). Though when you integrate then you should be careful, in general, with the integrations constants. Here they all turn out to be zero, hence all is excellent. $\endgroup$
    – Math-fun
    Mar 20, 2015 at 12:45
  • $\begingroup$ For the first integration it isn't zero.I had to put $x=0$ to find it. $\endgroup$ Mar 20, 2015 at 12:59
  • $\begingroup$ ups right! thanks! $\endgroup$
    – Math-fun
    Mar 20, 2015 at 13:45
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Since: $$-\frac{1}{k+1}+\frac{4}{k+2}-\frac{3}{k+3}=\frac{2k}{(k+1)(k+2)(k+3)}$$ we have: $$S_N=\sum_{n=1}^{N}\frac{k}{(k+1)(k+2)(k+3)}=\sum_{k=1}^{N}\left(\frac{-1/2}{k+1}+\frac{2}{k+2}+\frac{-3/2}{k+3}\right) $$ and since $(-1/2)+(2)+(-3/2)=0$ we have: $$ \lim_{N\to +\infty}S_N = \frac{-1/2}{1+1}+\frac{2}{1+2}+\frac{-1/2}{2+1}=\color{red}{\frac{1}{4}}.$$ The convergence of the original series is trivial since: $$0\leq \frac{k}{(k+1)(k+2)(k+3)}\leq\frac{1}{(k+2)(k+3)}$$ and: $$\sum_{k\geq 1}\frac{1}{(k+2)(k+3)}=\frac{1}{3}.$$

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Comparison:

$$\frac k{(k+1)(k+2)(k+3)}\le\frac{k+1}{(k+1)(k+2)(k+3)}=\frac1{k^2+5k+6}\le\frac1{k^2}$$

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  • $\begingroup$ Sorry and thanks. I forgot to say that the limit is also desired. Have edited. $\endgroup$
    – Yes
    Mar 20, 2015 at 11:58
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Observe that for each $k\in \mathbb{N}$, $\dfrac{k}{(k+1)(k+2)(k+3)}\leq \dfrac{k}{k.k.k}=\dfrac{1}{k^{2}}$ and $\sum\limits_{k=1}^{\infty}\dfrac{1}{k^{2}}$ converges.

Therefore $\sum\limits_{k=1}^{\infty}\dfrac{k}{(k+1)(k+2)(k+3)}$ converges by comparison test.

Also observe that $n\in \mathbb{N}$, $S_n=\sum\limits_{k=1}^{n}\dfrac{k}{(k+1)(k+2)(k+3)}=\sum\limits_{k=1}^{n}\left(-\dfrac{1}{2(k+1)}+\dfrac{2}{(k+2)}-\dfrac{3}{2(k+3)}\right)=\dots$

Therefore $\sum\limits_{k=1}^{\infty}\dfrac{k}{(k+1)(k+2)(k+3)}=\lim\limits_{n\to \infty}S_n=...$

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  • $\begingroup$ Thanks. I am aware of all that :) Just fail to see the limit. Have edited the question to specify this point. Sorry and thanks for your efforts. $\endgroup$
    – Yes
    Mar 20, 2015 at 12:01
  • $\begingroup$ Thanks. But is the second equality right? $\endgroup$
    – Yes
    Mar 20, 2015 at 12:15
  • $\begingroup$ I am convinced that the limit should be 1/4. I regret to say that something must go wrong in the reasoning above, which I believe to be the second equality. $\endgroup$
    – Yes
    Mar 20, 2015 at 12:27

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