Question

What is the interval of convergence of `sum_1^oo sin((pi*n)/2)/n^x`?

Answer 1

The series:

`sum_(n=0)^oo sin ((npi)/2)/n^x`

is convergent for `x in (0,+oo)`

Explanation:

We start by noting that:

`sin ((npi)/2) = {(0 " for " n = 2k ),((-1)^k " for " n = 2k+1):}`

with `k in NN_0`

So:

`sum_(n=0)^oo sin ((npi)/2)/n^x = sum_(k=0)^oo (-1)^k/(2k+1)^x = sum_(k=0)^oo (-1)^k/e^(xln(2k+1)) = sum_(k=0)^oo (-1)^ke^(-xln(2k+1))`

This is an alternating series so it is convergent if:

(i) `lim_(k->oo) e^(-xln(2k+1)) = 0`

(ii) `e^(-xln(2k+1)) > e^(-xln(2(k+1)+1))`

Both conditions are satisfied if the exponent is negative, and since `ln(2k+1) > 0` for `k > 0` this means that the series is convergent for `x > 0`.