Limit of Sequence: Determine m for Convergence

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Homework Statement

infƩn=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***
 
Last edited:
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Hi Tomp
[tex]\sum^{\infty}_{n=0}\frac{\cos (mn\pi)}{n+1}[/tex] is actually a series, not a sequence.
 
Last edited:
sharks said:
Hi Tomp
[tex]\sum^{\infty}_{n=0}\frac{\cos (mn\pi)}{n+1}[/tex] 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
 
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 [itex]m=\frac{1}{n\pi}[/itex]
 
sharks said:
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 [itex]m=\frac{1}{n\pi}[/itex]

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 :/