Finding the Limit of a Series - Solve Unknown Problem

[Frillth]

I have the following problem:
http://img432.imageshack.us/img432/9461/problemax9.jpg
I know for a fact that the answer is not 0, but I have no idea how to actually find the answer. I've never seen a similar problem before, and I'm not really sure how to start it.

 

[StatusX]

Can you find a function such that if you approximate its integral over some range by n rectangular strips, you get that sum? Then the limit would just be the integral over that range.

 

[quasar987]

I think there's a more direct approach.

Remember that a series is a sequence of partial sums. The general term of the sequence is

a_n=\sum_{k=1}^{n}\frac{k}{n^2} = \frac{1}{n^2}\sum_{k=1}^{n} k=?

You know how to do that sum on the left, and thus you can find an explicit form of the general term. Just take the limit.

 

[Frillth]

I'm not really sure what you mean by the explicit form of the general term. Can you give me a little more help, please?

 

[quasar987]

Do you know what

\sum_{k=1}^{n} k

sums to? It's sometimes called "Gauss' sum" named after the 8 years old who found the value of the sum in his head when asked to compute the sum of the first 50 integers.

 

[StatusX]

So you can get that sum directly in a few different ways, the simplest of which is to consider the pictures:

*0

**0
*00

***0
**00
*000

etc.

If you know that formula, that's definitely the easiest way. What I was suggesting might be overkill, and I was thinking of it because it's the way you would go about this if you had a higher power of k, ie:

\sum_{k=1}^{n}\frac{k^p}{n^{p+1}}

 

[Frillth]

Oh, duh! I can't believe I didn't think of that. Thanks!

 

[quasar987]

Generally, given an infinite series

\sum_{k=1}^{+\infty}b_n

you cannot put the general term a_n=\sum_{k=1}^n b_n of the sequence of partial sums in a friendly form that allows for direct computation of the value of the series by just taking the limit:

\lim_{n\rightarrow +\infty}a_n.

But the series you're dealing with is one of these rare case where the general term has a friendly form in terms of n that allows for this method of calculating the sum to work.