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Determine whether the following series is absolutely converger, conditionally convergent or divergent

$\sum\left(\frac{n+1}{2 n}\right)^{n} \cos n$

Apply B.C.T:

$|a n|=\left|\left(\frac{n+1}{2 n}\right)^{n}\right| \cdot|\cos (n)| =\left(\frac{n+1}{2 n}\right)^{n} \cdot|\cos (n)| \leq\left(\frac{n+1}{2 n}\right)$

$\cdot \sum\left(\frac{n+1}{n}\right)^{n}$ is +ve term Series.

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

$\lim _{n \rightarrow \infty} \sqrt[n]{b n}=\lim _{n \rightarrow \infty} \frac{n+1}{2 n}=\frac{\infty}{\infty} \Rightarrow$ Apply L'Hopitac Rule

$\lim _{n \rightarrow \infty} \sqrt[n]{ bn}=\lim _{n \rightarrow \infty} \frac{1}{2}=\frac{1}{2}<1 \Rightarrow \sum b n=\sum \left(\frac{n+1}{2 n}\right)^{n}$ is Convergent

$\sum |an| =\sum\left(\frac{n+1}{2 n}\right)^{n}\ |cos (n)|$ is convergent by B.c.T.

$\sum a n=\sum\left(\frac{n+1}{2 n}\right)^{n} \cdot \cos n$ is Absolutely convergent.

Determine whether the following series $A . C, \ C . C$ or Divergent

$\sum \frac{n^{2} \sin (3 n)}{(1.3)^{n}}$

$|a n|=\left|\frac{n^{2} \sin (3 n)}{(1.3)^{n}}\right|=\frac{n^{2}}{(1 \cdot 3)^{n}} \cdot|\sin (3 n)|$ Apply B.C.T $\sum | a n| \quad \sum | b n|$

$\frac{n^{2}}{(1.3)^{n}}|\sin (3 n)| \leq \frac{n^{2}}{(1 \cdot 3)^{n}}$

$-1 \leq \sin (3 n) \leq 1$

$b_{n}=\frac{n^{2}}{(1.3)^{n}} \rightarrow b_{n+1}=\frac{(n+1)^{2}}{(1 \cdot 3)^{n+1}}=\frac{(n+1)^{2}}{(1.3)^{1}(1.3)^{n}}$

$\lim _{n \rightarrow \infty} \frac{b n+1}{b n}=\lim _{n \rightarrow \infty} \frac{(n+1)^{2}}{(1.3)(1,3)^{n}} \cdot \frac{(1.3)^{n}}{n^{2}}= \frac{(n+1)^{2}}{1 \cdot 3 \cdot n^{2}}=\frac{n^{2}+2 n+1}{1.3 n^{2}}=\frac{\infty}{\infty}$

Apply L'Hopital Rule $=\lim _{n \rightarrow \infty} \frac{2 n+2}{1.3 \times 2 n}=\lim _{n \rightarrow \infty} \frac{2}{1-3 \times 2}=\lim _{n \rightarrow \infty} \frac{1}{1 \cdot 3}=\frac{1}{1.3}<1$

$\sum b n$ is convergent by Ratio test

$\sum|a n|$ is Convergent $by \ B \cdot C \cdot T$

$\sum a{n}$ is absolutely convergent.