Question

discuss the convergence of the series---
\[\large x+\frac{ 2^2x^2 }{ 2! }+\frac{ 3^3x^3 }{ 3!}+\frac{ 4^4x^4 }{ 4! }+......\infty\]

Answer 1

\[u_n=n^nx^n/n!\]
and
\[u_{n+1}=(n+1)^{n+1}*x^{n+1}/(n+1)!\]

Answer 2

\[\frac{ u_n }{ u_{n+1 }}=\frac{ n^n }{ (n+1)^n*x }=\frac{ 1 }{ [1+(1/n)^n] }*\frac{ 1 }{ x }\]

Answer 3

\[\lim_{n \rightarrow \infty}\frac{ u_n }{ u_{n+1} }=\frac{ 1 }{ ex }\]

Answer 4

thus by ratio test the series converges foe x<(1/e)
and diverges for x>(1/e)
but it fails for x=1/e
so ratio test is not working here
i dont know which test can be helpful :(

Answer 5

i think i should try raabe's or logarithmic test

Answer 6

hey it is working
\[\frac{ u_n }{u_{n+1} }=\frac{ e }{ (1+1/n)^n }\]
\[\log \frac{ u_n }{ u_{n+1} }=loge-nlog(1+1/n)=1-n(1/n-1/2n^2+1/3n^3+...)\]
so
\[\lim_{n \rightarrow \infty}n \log \frac{ u_n }{ u_{n+1} }=1/2<1\]
yes it worked!!!

Answer 7

converges for x<1/e
diverges for x>=(1/e)

Answer 8

\[\large S=\sum_{n=1}^\infty \frac{(nx)^n}{n!}\]
Using the ratio test, you should have
\[\large\begin{align*}\lim_{n\to\infty}\left|\frac{((n+1)x)^{n+1}}{(n+1)!}\cdot\frac{n!}{(nx)^n}\right|&=|x|\lim_{n\to\infty}\left|\frac{(n+1)^{n+1}x^n}{n+1}\cdot\frac{1}{(nx)^n}\right|\\\\
&=|x|\lim_{n\to\infty}\left|\frac{(n+1)^nx^n}{1}\cdot\frac{1}{(nx)^n}\right|\\\\
&=|x|\lim_{n\to\infty}\left(\frac{n+1}{n}\right)^n\\\\
&=e|x|\end{align*}\]
which tells you the series converges for \(|x|<\dfrac{1}{e}\) and diverges otherwise. I'm not sure why you thought you needed to do more than that?