# Sums and Limits

1. Feb 9, 2010

### Apteronotus

I've been trying to tackle a problem of the following form
$$lim_{n \rightarrow \infty} \sum_{k=0}^n f(k,x)$$

I know that the limit of each function is zero as n goes to infinity.
ie. $$lim_{n \rightarrow \infty} f(n,x) =0$$

But I'm not sure how to approach the problem above. I would greatly appreciate any thoughts/suggestions.

Thanks

2. Feb 9, 2010

### dacruick

what is f(k,x)

3. Feb 9, 2010

### Apteronotus

lol.... I didnt want to write it, thinking it may scare people off. Its a pretty complicated formula consisting of another summation.
But f(k,x) -> 0 as k->infinity.
The convergence to zero is governed by 1/sqrt(k).

4. Feb 9, 2010

### dacruick

well wouldnt the limit of the sum of f(k,x) as n goes to infinity exclusively consist of f(k,x)

5. Feb 9, 2010

### Landau

Basically we have the sum $$\sum_{k=0}^\infty a_k$$, where we know $$\lim_{k\to\infty}a_k=0$$. This information alone is not enough to conclude anything. For the sum to converge, it is necessary for a_k to converge to zero as k goes to infinity, but not sufficient. So if it would be the case that $$\lim_{k\to\infty}a_k\neq 0$$, then it follows that the sum does not converge. The fact that this condition is not sufficient, can be easily seen from the standard example $$\sum_{k=1}^\infty \frac{1}{k}$$, which diverges even though 1/k->0 if k->\infty.

You say "the convergence to zero is governed by 1/sqrt(k)". Could you be more precise?

We have $$\sum_{k=1}^\infty \frac{1}{\sqrt{k}}=\infty$$ since $$\frac{1}{\sqrt{k}}>\frac{1}{k}$$ for k>1, so probably your sum also diverges.