Limit of Sequence: Determine m for Convergence

[Tomp]

Homework Statement

^infƩ_n=0 cos(m*n*pi)/(n+1)
where m is a fixed integer. Determine the values of m, such that the series converges. Explain your reasoning in detail.

The attempt at a solution

I have figured out that cos(n*pi)/(n+1) can be represented as ((-1)^(n+1))/(n+1) (as it bounces back and forth from -1 to 1) and by the alternating series test, this converges.

However, I am unsure how to explain what m would do this.

I believe as m is an integer the cos(n*m*pi) term can't equal zero (need m = 1/2 and n=1 ect) so no matter what value m is, the sequence would converge, as the cos term can't be greater than -1 or 1? so m is any real integer (bit like n, though n is positive). Is this a bit trivial?
***And sorry, I understand this is a Series not a sequence***

 

[DryRun]

Hi Tomp
\sum^{\infty}_{n=0}\frac{\cos (mn\pi)}{n+1} is actually a series, not a sequence.

 

[Tomp]

Hi Tomp
\sum^{\infty}_{n=0}\frac{\cos (mn\pi)}{n+1} is actually a series, not a sequence.

yeah sorry, can't change the heading. hoping people would pick up on my correction in my edit

 

[DryRun]

You could prove that the limit has to be zero if the series converges, according to the nth-term test.

Try the squeeze theorem to show that the sequence is equal to zero, then solve for m.

I get m=\frac{1}{n\pi}

 

[Tomp]

You could prove that the limit has to be zero if the series converges, according to the nth-term test.

Try the squeeze theorem to show that the sequence is equal to zero, then solve for m.

I get m=\frac{1}{n\pi}

I have never learned about the squeeze theorem sorry. We have learned about the alternating, ration, comparison tests.

And m has to be an integer :/