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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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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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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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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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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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, $$\sum_{1}^{\infty}\dfrac{k}{(k+1)(k+2)(k+3)}=\frac14$$
$$\begin{align}\sum \limits_{k=1}^{\infty}\frac k{(k+1)(k+2)(k+3)}&= \sum \limits_{k=1}^{\infty}\dfrac{2}{k+2} - \dfrac{3}{2(k+3)} - \dfrac{1}{2(k+1)} \\~\\&=\frac{3}{2}\left(\sum \limits_{k=1}^{\infty}\dfrac{1}{k+2} -\dfrac{1}{k+3}\right) + \frac{1}{2}\left(\sum \limits_{k=1}^{\infty}\dfrac{1}{k+2}-\frac{1}{k+1}\right) \\~\\ &= \frac{3}{2}\left(\frac{1}{1+2}\right) + \frac{1}{2}\left(-\frac{1}{1+1}\right) \\~\\ &=\frac{1}{4}\end{align}$$