Examine the following series for convergence
∞Σn=1 [(n-1/(2n+3)]^n
Let an=(n−12n+3)na_n=(\dfrac{n-1}{2n+3})^nan=(2n+3n−1)n
Now, consider L=limn→∞an1/nL=\lim_{n\rightarrow \infty} {a_n}^{1/n}L=limn→∞an1/n
⇒L=limn→∞[(n−12n+3)n]1/n⇒L=limn→∞(n−12n+3)⇒L=limn→∞(1−1n2+3n)⇒L=12\Rightarrow L=\lim_{n\rightarrow \infty} [{(\dfrac{n-1}{2n+3})^n}]^{1/n} \\ \Rightarrow L=\lim_{n\rightarrow \infty} (\dfrac{n-1}{2n+3}) \\ \Rightarrow L=\lim_{n\rightarrow \infty} (\dfrac{1-\frac1n}{2+\frac3n}) \\ \Rightarrow L= \dfrac12⇒L=limn→∞[(2n+3n−1)n]1/n⇒L=limn→∞(2n+3n−1)⇒L=limn→∞(2+n31−n1)⇒L=21
Since, L<1L<1L<1, so by root test, Σan\Sigma a_nΣan converges.
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