Partial sums of $\sum_{n = 1}^{\infty}\frac{1}{n(n + 1)(n + 2)}$

Question

I have to calculate $\sum_{n = 1}^{\infty}\frac{1}{n(n + 1)(n + 2)}$. I decompose into partial sums and I get $$S_{m}=\sum_{n=1}^{m}\frac{-1}{x+1}+\sum_{n=1}^{m}\frac{1}{2(x+2)}+\sum_{n=1}^{m}\frac{1}{2x}$$ But I don't know how to proceed.

Answer 1

Your approach can work.

One way to start from your answer is to ask what is the total coefficient of $\frac1k$ in your sums, for $k=1,\dots,m+2.$

Turns out, for $k=3,\dots,m$ the sum is $\frac{1}{2}-1+\frac12=0.$

Then you need to deal with the additional cases $k=1,2,m+1,m+2.$

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Alternatively, note that your partial fractions can be written:

$$\begin{align}a_n&=\left(\frac{1}{2n}-\frac1{2(n+1)}\right)-\left(\frac{1}{2(n+1)}-\frac1{2(n+2)}\right)\\&=b_n-b_{n+1}\end{align}$$ where $$b_n=\frac{1}{2n}-\frac1{2(n+1)}$$

Answer 2

Hint: See that $$ \frac{2}{n(n + 1)(n + 2)} = \frac{(n+2) - n}{n(n + 1)(n + 2)} = \frac{1}{n(n+1)} - \frac{1}{(n+1)(n+2)}$$ and now the sum is telescopic!