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

1 Answer
Andrea S.
Mar 3, 2017

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.

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