What is the approach for solving problems involving sums and limits?

[Apteronotus]

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

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

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

Thanks

 

[dacruick]

what is f(k,x)

 

[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).

 

[dacruick]

well wouldn't the limit of the sum of f(k,x) as n goes to infinity exclusively consist of f(k,x)

 

[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}}&gt;\frac{1}{k} for k>1, so probably your sum also diverges.