Convergence of series <munderover> &#x2211;<!-- ∑ --> <mrow class="MJX-TeXAtom-ORD">

Patatiniuh

Patatiniuh

Answered question

2022-07-12

Convergence of series n = 1 ( e sin 1 n 1 )   x n , x > 0

Answer & Explanation

postojahob

postojahob

Beginner2022-07-13Added 13 answers

You have
lim n ( e sin ( 1 / n ) 1 ) x n x n / n = lim n n ( e sin ( 1 / n ) 1 ) = 1
since, if f ( x ) = e sin x , then f ( 0 ) = 1. So, since the series n = 1 x n n converges in (0,1) and diverges in [ 1 , ), so does your series.

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