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Test the convergence of the series $\sum_{n=1}^{\infty}\left(\frac{2 n+3}{3 n+2}\right)^{n}$

$(\cdots)^n$

Applying the root test:

$a n=\left(\frac{2 n+3}{3 n+2}\right)^{n} \ \Rightarrow \sqrt[n]{a n}=\sqrt[n]{\left(\frac{2 n+3}{3 n+2}\right)^{n}}=\frac{2 n+3}{3 n+2}$

$\lim _{n \rightarrow \infty}(\sqrt[n]{a n})=\lim_{n \rightarrow \infty} \left(\frac{2 n+3}{3 n+2}\right)=\frac{2(\infty)+3}{3(\infty)+2}=\frac{\infty}{\infty}$

use L'hopital Rule :

$\lim _{n \rightarrow \infty}(\sqrt[n]{a n})=\lim _{n \rightarrow \infty}\left(\frac{2 n+3}{3 n+2}\right)=\lim _{n \rightarrow \infty} \frac{2}{3}=\frac{2}{3}<1$

The series $\sum_{n=1}^{\infty}\left(\frac{2 n+3}{3 n+2}\right)^{n}$ is Convergent by Root test.

Test the series $\sum\left(1+\frac{2}{n}\right)^{n}$ for convergence

$(\ldots .)^{n}$ Root test

$a n=\left(1+\frac{2}{n}\right)^{n} \Rightarrow$ Apply the Root test $\sqrt[n]{a_{n}}=\sqrt[n]{\left(1+\frac{2}{n}\right)^{n}}=1+\frac{2}{n}$

$\lim _{n \rightarrow \infty}(\sqrt[n]{a n})=\lim _{n \rightarrow \infty}\left(1+\frac{2}{n}\right)=1+\frac{2}{\infty}=1+0=1$ Root test Fails.

lets try the test for Divargence or (nth term test) (NTT)

$\lim _{n \rightarrow \infty} a_{n}=\lim _{n \rightarrow \infty}\left(1+\frac{2}{n}\right)^{n}=\left(1+\frac{2}{\infty}\right)^{\infty}=1^{\infty}$

$y=\left(1+\frac{2}{n}\right)^{n} \Rightarrow \ln (y)=\ln \left(1+\frac{2}{n}\right)^{n}=n \ln \left(1+\frac{2}{n}\right)=\frac{\ln(1+\frac{2}{n})}{1/n}$

$\lim _{n \rightarrow \infty} \ln (y)=\lim _{n \rightarrow \infty}\left(\frac{\ln \left(1+\frac{2}{n}\right)}{1 / n}\right)=\frac{\ln (1)}{0}=\frac{0}{0}$ use L'hopital Rule

$\lim _{n \rightarrow \infty} \ln (y)=\lim _{n \rightarrow \infty} \frac{\frac{d}{d n}(\ln (1+2 / n))}{\frac{d}{d x}(1 / n)} =\lim _{n \rightarrow \infty} \frac{\frac{-2 / n^{2}}{1+2 / n}}{-1/n^2} =\lim _{n \rightarrow \infty} \frac{2}{1+2 / n}=2$

$\lim _{n \rightarrow \infty} e^{\ln (y)}=e^{2}$

$\lim _{n \rightarrow \infty} {an}=\lim _{n \rightarrow \infty}\left(1+\frac{2}{n}\right)^{n}=\lim _{n \rightarrow \infty} y=\lim _{n \rightarrow \infty} e^{\ln (y)}=e^{2} \neq 0$

$e^{\ln (y)}=y$

$\sum\left(1+\frac{2}{n}\right)^{n}$ is Divergent by the Test for Divergence.