Sums and Limits

  • #1
202
0
I've been trying to tackle a problem of the following form
[tex]
lim_{n \rightarrow \infty} \sum_{k=0}^n f(k,x)
[/tex]

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

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

Thanks
 

Answers and Replies

  • #2
1,033
1
what is f(k,x)
 
  • #3
202
0
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
1,033
1
well wouldnt the limit of the sum of f(k,x) as n goes to infinity exclusively consist of f(k,x)
 
  • #5
Landau
Science Advisor
905
0
Basically we have the sum [tex]\sum_{k=0}^\infty a_k[/tex], where we know [tex]\lim_{k\to\infty}a_k=0[/tex]. 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 [tex]\lim_{k\to\infty}a_k\neq 0[/tex], 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 [tex]\sum_{k=1}^\infty \frac{1}{k}[/tex], 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 [tex]\sum_{k=1}^\infty \frac{1}{\sqrt{k}}=\infty [/tex] since [tex]\frac{1}{\sqrt{k}}>\frac{1}{k}[/tex] for k>1, so probably your sum also diverges.
 

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