Q2004.
Can you help me with this question..
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n(n+1)}=\sum_{n=1}^{\i nfty}\frac{1}{n}-\frac{1}{n+1}=\sum_{n=1}^{\infty}\frac{1}{n}-\sum_{n=1}^{\infty}\frac{1}{n+1}=$
Now lets take out the first term of the series
$\displaystyle 1+\sum_{n=2}^{\infty}\frac{1}{n}-\sum_{n=1}^{\infty}\frac{1}{n+1}$
Now lets reindex the first sum with $\displaystyle k=n-1 \iff n=k+1$
Note: dont forget that the index is a "dummy variable"
$\displaystyle 1+\sum_{k=1}^{\infty}\frac{1}{k+1}-\sum_{n=1}^{\infty}\frac{1}{n+1}$
Since the index is a dummy vairable we can combine the sums let k=n
$\displaystyle 1+\sum_{n=1}^{\infty}\frac{1}{n+1}-\sum_{n=1}^{\infty}\frac{1}{n+1}=1+\sum_{n=1}^{\in fty}\left(\frac{1}{n+1}-\frac{1}{n-1} \right)=1+\sum_{n=1}^{\infty}0=1$