summation of difference

[carlacvds]

I need help showing \(\displaystyle \sum_{k=1}^{k=\infty}(\frac{1}{k}-\frac{1}{k+2})=\frac{3}{2}\)

I started out with writing out the terms: (1-1/3) + (1/2 - 1/5)
but this doesn't seem to be helping.

Can you point me in the right direction?
Thanks!

 

[pka]

\(\displaystyle \L\begin{array}{rcl}
S_4 & = & \sum\limits_{k = 1}^4 {\left( {\frac{1}{k} - \frac{1}{{k + 2}}} \right)} \\
& = & \left( {1 - \frac{1}{3}} \right) + \left( {\frac{1}{2} - \frac{1}{4}} \right) + \left( {\frac{1}{3} - \frac{1}{5}} \right) + \left( {\frac{1}{4} - \frac{1}{6}} \right) \\
& = & \left( {1 + \frac{1}{2}} \right) - \left( {\frac{1}{5} + \frac{1}{6}} \right) \\
\end{array}\)

\(\displaystyle \L\begin{array}{rcl}
S_n & = & \sum\limits_{k = 1}^n {\left( {\frac{1}{k} - \frac{1}{{k + 2}}} \right)} \\
& = & \left( {1 - \frac{1}{3}} \right) + \left( {\frac{1}{2} - \frac{1}{4}} \right) + \left( {\frac{1}{3} - \frac{1}{5}} \right) + \cdots + \left( {\frac{1}{n} - \frac{1}{{n + 2}}} \right) \\
& = & \left( {1 + \frac{1}{2}} \right) - \left( {\frac{1}{n+1} + \frac{1}{{n + 2}}} \right) \\
\end{array}\)

\(\displaystyle \L\lim _{n \to \infty } S_n = \frac{3}{2}\)

 

[carlacvds]

Thanks.

It looks like I made 2 mistakes. I had one of my terms wrong and
I stopped too early.

:lol: