Test the convergence of the series <munderover> &#x2211;<!-- ∑ --> <mrow class="MJX-TeXA

Thomas Hubbard

Thomas Hubbard

Answered question

2022-05-22

Test the convergence of the series
n = 1 n n 3 n n !

Answer & Explanation

bluayu0y

bluayu0y

Beginner2022-05-23Added 11 answers

You need to find the convergence of
n = 1 n n 3 n n !
So let u n = n n 3 n n ! which implies u n + 1 = ( n + 1 ) n + 1 3 n + 1 ( n + 1 ) !
Hence
u n u n + 1 = n n 3 n n ! . 3 n + 1 ( n + 1 ) ! ( n + 1 ) n + 1
u n u n + 1 = n n 3 n n ! . 3 n .   3   . ( n + 1 ) . n ! ( n + 1 ) n . ( n + 1 )
u n u n + 1 = 3. n n ( n + 1 ) n
u n u n + 1 = 3 ( 1 + 1 n ) n
lim n u n u n + 1 = 3 e > 1
(   S i n c e   lim n ( 1 + 1 n ) n = e )
Hence by D'Alemberts ratio test we have l > 1 ( l = lim n u n u n + 1 ) converges.

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