Q2004.

Can you help me with this question..

http://img401.imageshack.us/img401/8...09111306pm.png

Printable View

- Mar 3rd 2009, 08:46 AMchamprockseries..diverging, converging
Q2004.

Can you help me with this question..

http://img401.imageshack.us/img401/8...09111306pm.png - Mar 3rd 2009, 10:15 AMTheEmptySet
- Mar 3rd 2009, 10:25 AMmateoc15
This seems like a good candidate for "plug and chug". Just try a few values and see what happens. You'll figure it out.

- Mar 3rd 2009, 05:36 PMchamprock
- Mar 4th 2009, 03:58 PMTheEmptySet
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n(n+1)}=\sum_{n=1}^{\i nfty}\frac{1}{n}-\frac{1}{n+1}=\sum_{n=1}^{\infty}\frac{1}{n}-\sum_{n=1}^{\infty}\frac{1}{n+1}=$

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

$\displaystyle 1+\sum_{n=1}^{\infty}\frac{1}{n+1}-\sum_{n=1}^{\infty}\frac{1}{n+1}=1+\sum_{n=1}^{\in fty}\left(\frac{1}{n+1}-\frac{1}{n-1} \right)=1+\sum_{n=1}^{\infty}0=1$