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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.

 

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