1. Series Limit Problem

Consider SN=sum(n=1 to N) (1/(n^2+2n))

Write a formula for SN and find lim n-->infinity SN

So the terms are like 1/3+1/8+1/15+1/24.... essentially we are adding 5,7,9,11...etc..the odd numbers starting from 5 to the bottom. I don't know how to write a formula for this that is not recursive though.

I got SN=(SN-1)/(2N+1), but I think they want a formula that doesn't use recursion.

2. Originally Posted by zhupolongjoe
Consider SN=sum(n=1 to N) (1/(n^2+2n))

Write a formula for SN and find lim n-->infinity SN

So the terms are like 1/3+1/8+1/15+1/24.... essentially we are adding 5,7,9,11...etc..the odd numbers starting from 5 to the bottom. I don't know how to write a formula for this that is not recursive though.

I got SN=(SN-1)/(2N+1), but I think they want a formula that doesn't use recursion.
Note that $\frac{1}{n^2+2n} = \frac{1}{2} \left(\frac{1}{n} - \frac{1}{n+2} \right)$. Write out some of the terms and see what happens.

3. Originally Posted by zhupolongjoe
Consider SN=sum(n=1 to N) (1/(n^2+2n))

Write a formula for SN and find lim n-->infinity SN

So the terms are like 1/3+1/8+1/15+1/24.... essentially we are adding 5,7,9,11...etc..the odd numbers starting from 5 to the bottom. I don't know how to write a formula for this that is not recursive though.

I got SN=(SN-1)/(2N+1), but I think they want a formula that doesn't use recursion.

Elephant hint: $\frac{1}{n(n+2)}=\frac{1}{2}\,\left(\frac{1}{n}-\frac{1}{n+2}\right)$

Tonio

4. Thanks, is it 3/4?

5. Originally Posted by zhupolongjoe
Thanks, is it 3/4?

Yup

Tonio