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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.

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