Answer to Question #346386 in Real Analysis for Mr Alex

Question #346386

Examine the following series for convergence: n=0(n22n+3)n\sum _{n=0}^{\infty }\left(\frac{n-2}{2n+3}\right)^n


1
Expert's answer
2022-05-31T12:23:00-0400

Use the Root Test


limnann=limn(n22n+3)nn=limnn22n+3\lim\limits_{n\to \infin}\sqrt[n]{|a_n|}=\lim\limits_{n\to \infin}\sqrt[n]{|(\dfrac{n-2}{2n+3})^n|}=\lim\limits_{n\to \infin}|\dfrac{n-2}{2n+3}|

=limnn/n2/n2n/n+3/n=limn12/n2+3/n=\lim\limits_{n\to \infin}|\dfrac{n/n-2/n}{2n/n+3/n}|=\lim\limits_{n\to \infin}|\dfrac{1-2/n}{2+3/n}|

=102+0=12<1=|\dfrac{1-0}{2+0}|=\dfrac{1}{2}<1

Thus the given series converges by the Root Test.


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