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

Answer 1

Cameron Alexander

The series:

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

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

To begin, we observe 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)) #

Since this is an alternating series, convergence occurs when:

(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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Answer 2

Cameron Alexander

The interval of convergence for the series ( \sum_{n=1}^\infty \frac{\sin\left(\frac{\pi n}{2}\right)}{n^x} ) is ( x > 0 ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.