Does the sum of $\sum_{n=1}^{\infty}{\frac{2}{n^2+2n}}$ depend at which $n$ the series starts?

Question

I want to find the sum of the following series:

$$\sum_{n=1}^{\infty}{\frac{2}{n^2+2n}}$$

Now, this should be a telescopic series. It is equal to

$$2\sum_{n=1}^{\infty}{\frac{1}{n^2+2n}=2\sum_{n=1}^{\infty}{\frac{1}{n(n+2)}}=2\sum_{n=1}^{\infty}\frac{1}{2}{\left[{\frac{1}{n}-\frac{1}{n+2}}\right]}}=\sum_{n=1}^{\infty}{\left(\frac{1}{n}-\frac{1}{n+2}\right)}$$

This series is in the same form of

$$\sum_{n=m}^{\infty}\left(a_n-a_{n+k}\right)$$

In fact, $a_n=\frac 1 n$ and $a_{n+2}=\frac{1}{n+2}$. Since $k=2$ the series has this sum:

$$a_1+a_2=1+\frac{1}{2}=\frac{3}{2}$$

Now, my question is if it's safe to say that in a telescopic series it does not matter at which $n$ the series "starts". To find the sum we haven't considered $n=1$. It could have been any number. Any hints?

Answer 1

Since

$$\sum_{n=1}^{\infty}{\frac{2}{n^2+2n}}=S\le 2\sum_{n=1}^{\infty}{\frac{1}{n^2}}=\frac{\pi^2}3$$

we have that

$$\sum_{n=k}^{\infty}{\frac{2}{n^2+2n}}=S-\sum_{n=1}^{k-1}{\frac{2}{n^2+2n}}$$

Answer 2

Consider a telescopic series, (This one is easier to understand.)

$$\sum_{n=m}^{\infty}{\frac{1}{n}-\frac{1}{1+n}}$$

$$=\frac{1}{m}$$

Wouldn't it depend on what $'m'$ is?

Answer 3

It doesn't matter in the sense that:

1. If the sum converges, then it converges for any starting index
2. You can always transform the expression as a sum or difference of the sum starting from zero, plus or minus the missing terms.

Here, you can just see that some terms telescope when you compute the partial sum. Taking the limit, the term of high order cancels out and you get the right result.