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###### ${selected_topic_name} • Notes • Comments & Questions Test the convergence of $\sum_{n=1}^{\infty} 3(4)^{1-n}(3)^{n-2}$ and find its sum if it converges. $a n=3(4)^{1-n}(3)^{n-2}=3 \cdot \frac{4^{1}}{4^{n}} \cdot \frac{3^{n}}{3^{2}}=\frac{4}{3} \cdot\left(\frac{3}{4}\right)^{n}$ $\sum_{n=1}^{\infty} 3 \cdot(4)^{1-n}(3)^{n-2}=\frac{4}{3} \cdot\left(\frac{3}{4}\right)^{n}$ Geometric series with $r=\frac{3}{4}<1$ Convergent series $a=\frac{4}{3} \cdot\left(\frac{3}{4}\right)^{n}=\frac{4}{3} \cdot\left(\frac{3}{4}\right)^1=1$ $Sum=\frac{a}{1-r}=\frac{1}{1-3 / 4}=4$ Let $x=0 . \overline{23}$ express x as the sum of a geometric series and find the integrals m & n such that $x=\frac{m}{n}$ $x=0 . \overline{23}=0.232323 \dots$ $=0.23+0.0023+0.000023 \cdots$ $=23\left(10^{-2}\right)+23\left(10^{-4}\right)+23\left(10^{-6}\right) \dots$ $=23\left(10^{-2}\right)^{1}+23\left(10^{-2}\right)^{2}+23\left(10^{-2}\right)^{3} \cdots$ $x=\sum_{n=1}^{\infty} 23\left(10^{-2}\right)^{n}=\sum_{n=1}^{\infty} \frac{23}{\left(10^{2}\right)^{n}}=\sum_{n=1}^{\infty} \frac{23(1)}{(100)^{n}}$ $x=\sum_{n=1}^{\infty} 23 \cdot\left(\frac{1}{100}\right)^{n} \Rightarrow r=\frac{1}{100}<1$ Convergent series $a=23 \cdot\left(\frac{1}{100}\right)^{1}=\frac{23}{100}$ $x= {sum}=\frac{a}{1-r}=\frac{23 / 100}{1-1 / 100}=\frac{23}{100-1}=\frac{23}{99}$ $x=\frac{23= m}{99= n}$ m = 23 n = 99 Test the convergence $\sum_{n=1}^{\infty} \ln \left(\frac{7 n+1}{2 n+3}\right)$ $a n=\ln \left(\frac{7 n+1}{2 n+3}\right)$ $\lim _{n \rightarrow \infty} a n=\lim _{n \rightarrow \infty}\ln\left (\frac{7 n+1}{2 n+3}\right)=\frac{\infty}{\infty}$ $\ln \lim _{n \rightarrow \infty}\left(\frac{7 n+1}{2 n+3}\right)=\ln \lim _{n \rightarrow \infty}\left(\frac{7 x+1}{2 x+3}\right)=\ln \lim _{n \rightarrow \infty}\left(\frac{7}{2}\right)$ $\ln \lim _{n \rightarrow \infty}\left(\frac{7 n+1}{2 n+3}\right)=\ln \lim _{n \rightarrow \infty}\left(\frac{7}{2}\right)=\ln \left(\frac{7}{2}\right)$ $\lim _{n \rightarrow \infty} a_{n}=\frac{7}{2} \neq 0$ Divergent test for divergence or by Nth term test (N.T.T) Test for convergence$$\sum_{n=1}^{\infty} \sqrt[3]{\frac{2 n+5}{16 n-3}}$

$$a n=\sqrt[3]{\frac{2 n+5}{16 n-3}}$$

$\lim _{n \rightarrow \infty} a n=\lim _{n \rightarrow \infty} \sqrt[3]{\frac{2 n+5}{16 n-3}}=\sqrt[3]{\lim _{n \rightarrow \infty} \frac{2 n+5}{16 n-3}}=\frac{\infty}{\infty}$

$=\sqrt[3]{\lim _{n \rightarrow \infty} \frac{2}{16}}=\sqrt[3]{\frac{1}{8}}=\frac{1}{2} \neq 0$

The series is Divergent by N.T.T

or Test is Divergence