Summation of sin(pi*n/2)/2: Is the Execution Correct?

[mr-feeno]

Homework Statement

\sum_{n=2}^{\infty}sin(\frac{\pi*n}{2})/{2}I don't have a solution, and wondered if the execution is correct.

The Attempt at a Solution

I thought that one can use comparison test where; \sum b_n= \frac{1}{n^{1/2}}.
Since p<1 ---> divergent. But many of the students says it converges. Some suggestions?

We know that the series is alternating, and if I use its test I get that it converges. Reckon that perhaps the fault lies there?

 

[hilbert2]

Let's define a function that's related to that series:

$f(x) = \frac{1}{2}\sum_{n=2}^{\infty}sin(\frac{n\pi x}{2})$ .

Now obviously the sum of the original series, if it exists, is $f(1)$. Is the expression of $f(x)$ the Fourier series of some function that you know?

EDIT1: This might help: http://functions.wolfram.com/GeneralizedFunctions/DiracDelta/06/01/

Is the Dirac delta an acceptable function in the sense of rigorous mathematics?

EDIT2: Also, if a sum of terms $a_k$ is convergent, what can we tell about the limit of the sequence $(a_k)$ when $k \rightarrow \infty$ ?

 

[Ray Vickson]

Homework Statement

\sum_{n=2}^{\infty}sin(\frac{\pi*n}{2})/{2}I don't have a solution, and wondered if the execution is correct.

The Attempt at a Solution

I thought that one can use comparison test where; \sum b_n= \frac{1}{n^{1/2}}.
Since p<1 ---> divergent. But many of the students says it converges. Some suggestions?

We know that the series is alternating, and if I use its test I get that it converges. Reckon that perhaps the fault lies there?

Well, the actual terms for $n = 2,3,4, \ldots$ are 0, -1/2, 0, 1/2, 0, -1/2, 0, 1/2, 0, ... . Do you think those terms give a convergent series?

 

[hilbert2]

Maybe in the sense of Cesaro summability...

 

[Ray Vickson]

Maybe in the sense of Cesaro summability...

I would rather that the OP have a good grasp of ordinary convergence/divergence before exploring more arcane topics like cesaro or Abel or ... summability.