Question

Check the convergence$\sum _{k=1}^{\infty }\frac{k!{10}^k}{3^k}$

Short answer

Converges

Step-by-step solution

Step 1. Given

The given series is$\sum _{k=1}^{\infty }\frac{k!{10}^k}{3^k}$

Step 2. Ratio test

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

Step 3. Checking the convergence 

$$\begin{gathered} \mathrm{Now},\mathrm{calculate}\mathrm{the}\mathrm{value}\mathrm{of}\frac{a_{k+1}}{a_k} \\ \mathrm{Consider}\mathrm{the}\mathrm{general}\mathrm{term}as=\frac{k!{10}^k}{3^k}...(1 \\ \mathrm{Therefore},\mathrm{by}\mathrm{substituting}\mathrm{k}=\mathrm{k}+1\ln (1) \\ {a}_{k+1}=\frac{(k+1)!{10}^{(k+1)}}{3^{k+1}} \\ Now,\frac{a_{k+1}}{a_k}=\frac{{\displaystyle \frac{{\displaystyle (k+1)!{10}^{(k+1)}}}{{\displaystyle 3^{k+1}}}}}{\frac{\left(k\right)!{10}^{\left(k\right)}}{3^k}} \\ U\sin gformulan!=n(n-1)! \\ \frac{a_{k+1}}{a_k}=\frac{10(k+1)}{3} \\ Takinglimits \\ \underset{k\to \infty }{\lim }\frac{a_{k+1}}{a_k}=\underset{k\to \infty }{\lim }\frac{10(k+1)}{3} \\ L=\infty \\ Theseriesconverges \end{gathered}$$