Use the Root Test to determine the convergence or divergence of the series. sum_{n=1}^infty(frac{3n+2}{n+3})^n

babeeb0oL

babeeb0oL

Answered question

2021-02-21

Use the Root Test to determine the convergence or divergence of the series.
n=1(3n+2n+3)n

Answer & Explanation

doplovif

doplovif

Skilled2021-02-22Added 71 answers

The given series is n=1(3n+2n+3)n
Ratio test:
The series n=1an converges if limn|an|n<1
The series n=1an diverges if limn|an|n>1
In the given series n=1(3n+2n+3)n,an=(3n+2n+3)n
Find the limit limn|an|n
limn|an|n=limn|(3n+2n+3)n|1n
=limn(3n+2n+3)
=limn(3+2n1+3n)
=3+21+3
=3+01+0
=3
>1
By the Ratio test, the series n=1(3n+2n+3)n diverges.

Jeffrey Jordon

Jeffrey Jordon

Expert2021-12-27Added 2605 answers

Answer is given below (on video)

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