Question

Check the convergence$\sum _{k=0}^{\infty }\frac{5^{2k+1}}{(2k+1)!}$

Short answer

The series converges

Step-by-step solution

Step 1. Given

The given series is$\sum _{k=0}^{\infty }\frac{5^{2k+1}}{(2k+1)!}$.

Step 2.  Ratio test

$$\begin{gathered} \mathrm{According}\mathrm{to}\mathrm{the}\mathrm{RatioTest},\underset{\mathrm{k}=1}{\overset{\infty }{\sum {\mathrm{a}}_{\mathrm{k}}}}\mathrm{be}\mathrm{the}\mathrm{series}\mathrm{with}\mathrm{positive}\mathrm{terms},\mathrm{if}\mathrm{L}=\underset{\mathrm{k}\to \infty }{\lim }\frac{{\mathrm{a}}_{\mathrm{k}+1}}{{\mathrm{a}}_{\mathrm{k}}} \\ 1.\mathrm{If}L<1\mathrm{series}\mathrm{converges}. \\ 2.\mathrm{If}\mathrm{L}>1\mathrm{series}\mathrm{diverges}. \\ 3.\mathrm{If}\mathrm{L}=1\mathrm{the}\mathrm{test}\mathrm{is}\mathrm{inconclusive}. \end{gathered}$$

Step 3. Checking the convergence

$$\begin{gathered} \mathrm{Now},\mathrm{calculate}\mathrm{the}\mathrm{value}\mathrm{of}\frac{a_{k+1}}{a_k} \\ \mathrm{Consider}\mathrm{the}\mathrm{general}\mathrm{term}as=\frac{5^{2k+1}}{(2k+1)!.}...(1 \\ Therefore,bysubstituting\mathrm{k}=\mathrm{k}+1\ln (1) \\ {a}_{k+1}=\frac{5^{2(k+1)+1}}{\left(2\right(k+1)+1)!} \\ =5^{2k+3(}2k+3)! \\ Now,\frac{a_{k+1}}{a_k}=\frac{{\displaystyle \frac{5^{2k+3}}{(2k+3)}}}{5^{2k+!}} \\ U\sin gformulan!=n(n-1)! \\ \frac{a_{k+1}}{a_k}=\frac{25}{(2k+3)(2k+2)} \\ Takinglimits \\ \underset{k\to \infty }{\lim }\frac{a_{k+1}}{a_k}=\underset{k\to \infty }{\lim }\left(\frac{25}{k\left(2+{\displaystyle \frac{3}{k}}\right)\left(2+{\displaystyle \frac{2}{k}}\right)}\right) \\ L=0 \end{gathered}$$