Sum $\sum_{n=1}^{\infty} \frac{1}{n}-\frac{1}{n+2}$ I have to first prove whether the following series is convergent, and then find its limit if it exists:
$\sum_{n=1}^{\infty} \frac{1}{n}-\frac{1}{n+2}$
So, I have proved that it is convergent, but I'm having trouble in finding the limit. Can someone help me out?
 A: If the problem is due to 
$$
\frac{1}
{n} - \frac{1}
{{n + 2}}
$$ 
write it as 
$$
a_n  = \left( {\frac{1}
{n} - \frac{1}
{{n + 1}}} \right) + \left( {\frac{1}
{{n + 1}} - \frac{1}
{{n + 2}}} \right)
$$
At this point both the terms in the brackets telescoping. So I thin now it is easy.
A: $\sum_{n=1}^\infty (\frac 1n -\frac 1{n+2}) = $ (if it exists at all)
$\lim_{M\to \infty}(\sum_{n=1}^M \frac 1n -\frac 1{n+2})$ and
$\sum_{n=1}^M (\frac 1n -\frac 1{n+2}) = \sum_{n=1}^M \frac 1n - \sum_{n=1}^M\frac 1{n+2} = $
$\sum_{n=1}^m\frac 1n -\sum_{n=3}^{M+2} \frac 1n =$
$ 1 + \frac 12 - \frac 1{M+1} - \frac 1{M+2}$
So $\sum_{n=1}^\infty (\frac 1n -\frac 1{n+2}) = \lim_{M\to \infty}(1 + \frac 12 - \frac 1{M+1} - \frac 1{M+2})$ which....
I'll leave to you.
A: The ans is $\frac{3}{2}$. Compute the sequence of partial sum and you'll figure it out.
Spoiler :
$S_n = (1+\frac{1}{2}+...+\frac{1}{n})-(\frac{1}{3}+\frac{1}{4}+...+\frac{1}{n+2})$
$= (1+\frac{1}{2}+...+\frac{1}{n})+(1+\frac{1}{2})-(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{n+2})$
$=(\gamma_{n} + ln {n}) + \frac{3}{2} - (\gamma_{n+2} + ln (n+2))$
$=(\gamma_n - \gamma_{n+2}) +ln {\frac{n}{n+2}}+ \frac{3}{2}$
Therefore $\lim_{n\to \infty} S_n = \frac{3}{2}$
