Question #291363

Examine the following series for convergence


∞Σn=1 [(n-1/(2n+3)]^n

1
Expert's answer
2022-01-31T15:02:34-0500

Solution:

Let an=(n12n+3)na_n=(\dfrac{n-1}{2n+3})^n

Now, consider L=limnan1/nL=\lim_{n\rightarrow \infty} {a_n}^{1/n}

L=limn[(n12n+3)n]1/nL=limn(n12n+3)L=limn(11n2+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

Since, L<1L<1, so by root test, Σan\Sigma a_n converges.


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