Cos(2n)/n^0.5 convergent or not?

[lyranger]

How to determine if cos(2n)/n^0.5 is convergent?

I've tried summation by parts and other methods but none of them works.

Hope someone can help me. Cheers

Sorry its the sum where n goes from 1 to infinity

 

[Sourabh N]

For what value of n?
For n going to 0, denominator goes to 0, numerator to 1, so no, not convergent.
For n going to infinity, denominator goes to infinity, numerator oscillates, so no, not convergent.

 

[lyranger]

For what value of n?
For n going to 0, denominator goes to 0, numerator to 1, so no, not convergent.
For n going to infinity, denominator goes to infinity, numerator oscillates, so no, not convergent.

Sorry it's the sum of cos(2n)/n^0.5 for n from 1 to infinity

 

[sweet springs]

Hi.

In Excel sheet the sum stays around -0.5 but does not seem to converge in Excel up to n=1000.

Regards.

 

[lyranger]

zz i was trying to prove it. but anyway thanks a lot for all the effort with excel mate

 

[uart]

Hi lyranger, this is just an outline, but I think the key is to break it up into (a sum of) partial sums, with the partials sums (indexed by say k) running over $(k \pi - \pi/2) \leq 2n < (k \pi + \pi/2)$.

Although the actual values of those partial sums isn't particularly tractable, at least we know that they are strictly alternating and we can easily put bounds on their absolute values. For example we can show that the number of terms in each partial sum is at most two and therefore the maximum absolute value of the k_th partial sum is ...

Hopefully that will put you on the right track. The idea is that we can ultimately use the alternating series theorem.

 

[Ray Vickson]

How to determine if cos(2n)/n^0.5 is convergent?

I've tried summation by parts and other methods but none of them works.

Hope someone can help me. Cheers

Sorry its the sum where n goes from 1 to infinity

"its" should be "it's" = "it is".

If we replace the sum by an integral, it converges. Maple gives
$$\int_{1}^{\infty} \frac{\cos(2x)}{\sqrt{x}}\, dx = \frac{\sqrt{\pi}}{2}-\sqrt{\pi}\text{FresnelC}\left(\frac{2}{\sqrt{\pi}}\right) \doteq - 0.4489667705.$$
I am not sure whether or not this helps when looking at the sum instead of the integral.

However, numerically it seems to converge. Defining S(N) = sum[cos(2n)/sqrt(n),n=1..N), Maple gives S(1000) = -0.5163020976, S(2000) = -0.5330025718, S(3000) = -0.5141927050, etc.

RGV