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$$\sum_{n=0}^{\infty } \left ( \frac{1}{n.(n+1).(n+2)))} \right )$$

I've tried all of the tests but I couldnt solve this.

Does it converges? What is the sum if it converges?

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Note your series can' start at $n=0$. This fixed, yes, this converges by comparison with $\sum 1/n^3$. To compute the sum, try partial fraction decomposition. Or even shorter, the partial partial fraction decomposition of Andre Nicolas. – 1015 Feb 19 '13 at 23:01

If $(u_k)$ is a sequence, we have, by a simple change of index, for forall $n\in\mathbb{N}^*$: $$\sum_{k=1}^n( u_{k}-u_{k+1})=u_1-u_{n+1}.$$ With $$\frac{1}{n(n+1)}-\frac{1}{(n+1)(n+2)}=\frac{2}{n(n+1)(n+2)}$$ and if we pose $u_n= \frac{1}{n(n+1)}$, we find

$$\sum_{n=1}^{+\infty}\frac{1}{n(n+1)(n+2)}=\lim_n \frac{1}{2}\sum_{k=1}^n (u_k-u_{k+1})=\lim_n \frac{1}{2}(u_1-u_{n+1})=\frac{u_1}{2}=\frac{1}{4}.$$

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You want to start at $n=1$, in order to avoid division by $0$.

Hint: $$\dfrac{1}{n(n+1)}-\dfrac{1}{(n+1)(n+2)}=\dfrac{2}{n(n+1)(n+2)}.$$

You should be able to use this to get an explicit formula for the sum of the first $n$ terms of your series.

If you are only interested in proving convergence, note that the $n$-th term is $\lt \dfrac{1}{n^3}$. You may have already proved that $\sum \frac{1}{n^3}$ converges, say using the Integral Test. Then your series converges by Comparison.

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