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• Notes

Ratio and Root test:

Ratio test:  $\lim _{n \rightarrow \infty}\left(\frac{a_{n+1}}{a_{n}}\right)=L$

Root test: $\lim _{n \rightarrow \infty} \sqrt[n]{a_ n}=L$

Ratio test: $n^{a} \cdot a^{n}, n^{n}, n !$

Root test : $(\ldots \ldots)^{n}$

$n !=n(n-1)(n-2) \dots 1$

$5 !=5 \times 4 \times 3 \times 2 \times 1$

Test the convergence of the series $\sum \frac{2^{n}}{n !}$

$2^{n}, n !$

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

$a n=\frac{2^{n}}{n !} \quad , a_{n+1}=\frac{2^{n+1}}{(n+1) !}, \frac{a n+1}{a_{n}}=\frac{\frac{2^{n+1}}{(n+1) !}}{\frac{2^{n}}{n !}}=\frac{2^{n+1}}{(n+1) !} \times \frac{n !}{2^{n}}$

$\frac{a_{n+1}}{a n}=\frac{\left(2^{n+1}\right) n !}{2^{n} \times(n+1) !}$    but $(n+1) !=(n+1)(n)(n-1)\cdots$

$=(n+1) n !$

$\frac{a_{n+1}}{a_{n}}= \frac{(2^{n+1})n!}{2^n\times (n+1)n!}=\frac{2}{n+1}$

$\lim _{n \rightarrow \infty}\left(\frac{a_{n+1}}{a n}\right) =\lim _{n \rightarrow \infty}\left(\frac{2}{n+1}\right) =\frac{2}{\infty+1}=\frac{2}{\infty}=0<1$ Convergent

Series by Ratio test

Test the series $\sum \frac{(n !)^{2}}{(2 n) !}$

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

$a_{n}=\frac{(n!)^{2}}{(2 n) !} , \quad a_{n+1}=\frac{((n+1) !)^{2}}{(2(n+1)) !}=\frac{((n+1) !)^{2}}{(2 n+2) !}$

$\frac{a_{n+1}}{a n}=\frac{((n+1) !)^{2}}{(2 n+2) !} \times \frac{(2 n) !}{(n !)^{2}} =\frac{\left((n+1)(n)!)^{x}\right.}{(2 n+2)(2 n+1)(2 n) !} \times \frac{(2 n) !}{(n!)^{2}}$

$\frac{a_{n}+1}{a_{n}}= \frac{(n+1)^{2}}{(2 n+2)(2 n+1)} =\frac{(n+1)^{2}}{2(n+1)(2 n+1)} =\frac{n+1}{2(2 n+1)}$

$\lim _{n \rightarrow \infty}\left(\frac{a n+1}{a n}\right)=\lim _{n \rightarrow \infty} \frac{n+1}{2(2 n+1)}=\frac{\infty}{\infty}$                Apply L'Hopital Rule

$\lim _{n \rightarrow \infty} \frac{1}{4}=\frac{1}{4}<1 \sum \frac{(n !)^{2}}{(2 n) !}$   is Convergent by Ratio test.