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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