# Prove that the series $\sum_{1}^{\infty}\frac{k}{(k+1)(k+2)(k+3)}$ converges and find its limit

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

• 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 ? Mar 20, 2015 at 12:11
• Thanks. Seems reasonable, I am trying :)
– Yes
Mar 20, 2015 at 12:13

\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}

• That is the right track.
– Yes
Mar 20, 2015 at 12:16
• Haha it was a fun telescoping sum $$\sum\limits_{r=1}^{\infty}f(r) - f(r+1) = f(1)$$ provided $\lim\limits_{r\to\infty }f(r) = 0$ Mar 20, 2015 at 12:18
• Nice. I should know better than to give subtle hints. :) Mar 20, 2015 at 12:21

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.

• 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.
– Yes
Mar 20, 2015 at 12:24

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, $$\sum_{1}^{\infty}\dfrac{k}{(k+1)(k+2)(k+3)}=\frac14$$

• 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. Mar 20, 2015 at 12:45
• For the first integration it isn't zero.I had to put $x=0$ to find it. Mar 20, 2015 at 12:59
• ups right! thanks! Mar 20, 2015 at 13:45

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: $$\sum_{k\geq 1}\frac{1}{(k+2)(k+3)}=\frac{1}{3}.$$

Comparison:

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

• Sorry and thanks. I forgot to say that the limit is also desired. Have edited.
– Yes
Mar 20, 2015 at 11:58

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

• 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.
– Yes
Mar 20, 2015 at 12:01
• Thanks. But is the second equality right?
– Yes
Mar 20, 2015 at 12:15
• 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.
– Yes
Mar 20, 2015 at 12:27