# Convergence or not

## Homework Statement

$$\sum_{n=2}^{\infty}sin(\frac{\pi*n}{2})/{2}$$

I dont 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
Gold Member
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:
Ray Vickson
Homework Helper
Dearly Missed

## Homework Statement

$$\sum_{n=2}^{\infty}sin(\frac{\pi*n}{2})/{2}$$

I dont 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?

Ray Vickson