1. ## series..diverging, converging

Q2004.
Can you help me with this question..

2. Originally Posted by champrock
Q2004.
Can you help me with this question..
Note that $\displaystyle \frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$

so this telescopes to 1. So it is none of these.

3. This seems like a good candidate for "plug and chug". Just try a few values and see what happens. You'll figure it out.

4. Originally Posted by TheEmptySet
Note that $\displaystyle \frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$

so this telescopes to 1. So it is none of these.
i has used your method and found the value to be 1-Lim(1/n+1) where n tends to infinity.

So this becomes 1-(some minutest number close to 0) . so shouldnt the answer be "some number between 0 and 1" ?

i am confused on this part.

5. Originally Posted by champrock
i has used your method and found the value to be 1-Lim(1/n+1) where n tends to infinity.

So this becomes 1-(some minutest number close to 0) . so shouldnt the answer be "some number between 0 and 1" ?

i am confused on this part.
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n(n+1)}=\sum_{n=1}^{\i nfty}\frac{1}{n}-\frac{1}{n+1}=\sum_{n=1}^{\infty}\frac{1}{n}-\sum_{n=1}^{\infty}\frac{1}{n+1}=$

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

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