How to determine that this series is conditionally convergent? <munderover> &#x2211;<!-- ∑ -

Jaycee Mathis

Jaycee Mathis

Answered question

2022-05-22

How to determine that this series is conditionally convergent?
n = 1 n n + 1 sin π x x p + 1 d x , 0 < p 1

Answer & Explanation

Mya Hurst

Mya Hurst

Beginner2022-05-23Added 13 answers

Notice that sin ( π x ) doesn't change its sign on the interval [ n , n + 1 ] since its zeros are x = k Z . By the mean value theorem there's a c n [ n , n + 1 ] such that
n n + 1 sin ( π x ) x p + 1 d x = 1 ( c n ) p + 1 n n + 1 sin ( π x ) d x
For the last integral, we have
n n + 1 sin ( π x ) d x = 1 π cos ( π x ) | n n + 1 = 1 π ( cos ( ( n + 1 ) π ) cos ( n π ) ) = 2 ( 1 ) n π
so you have to establish the convergence of
2 π n = 1 ( 1 ) n ( c n ) p + 1
which follows from the alternating series test, and the divergence of
2 π n = 1 1 ( c n ) p + 1
which follows from a comparison with the divergent series
2 π n = 1 1 ( n + 1 ) p + 1 = 2 π n = 2 1 n p + 1

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?