# Convergence or not

1. Dec 4, 2016

### mr-feeno

1. The problem statement, all variables and given/known data
$$\sum_{n=2}^{\infty}sin(\frac{\pi*n}{2})/{2}$$

I dont have a solution, and wondered if the execution is correct.

3. 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?

2. Dec 4, 2016

### 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 $k \rightarrow \infty$ ?

Last edited: Dec 4, 2016
3. Dec 4, 2016

### Ray Vickson

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?

4. Dec 4, 2016

5. Dec 4, 2016

### Ray Vickson

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.