Test the convergence of the series $$ x+\frac{2^{2} x^{2}}{2 !}+\frac{3^{3} x^{3}}{3 !}+\frac{4^{4} x^{4}}{4 !}+\cdots $$ Assume $x>0$ and examine all possibilities.
My approach:
Using ratio test I am getting
$$ L=\lim _{n \rightarrow \infty}\left|\frac{a_{n+1}}{a_{n}}\right| = |e.x| >1 $$ Can I conclude the series is divergent?
On
Careful : you have to check for $x>0$ what is the limit of $\left|\dfrac{a_{n+1}x^{n+1}}{a_nx^n}\right| = \left|\dfrac{a_{n+1}}{a_n}x\right|$. This is equal to $\left|\dfrac{(n+1)^{n+1}}{n^n(n+1)}x \right|$ with limit $|ex|$ Thus the limit is $<1$ if and only if $x < 1/e$. Then the radius of convergence is $1/e$. You have to check manually for $|x|= 1/e$ if it converges.
For x = 1/e ; U can go for a ratio test and by then u will find the series will come as a divergent one $c=\lim _{n \rightarrow \infty} \frac{b_{n}}{a_{n}}$ Let, $b_{n}=\frac{1}{\sqrt{n}}=\frac{1}{n^{1 / 2}}$ So, $\lim _{n \rightarrow \infty}$ $ \frac{b n}{a_{n}}=\frac{\frac{1}{\sqrt{n}}}{\frac{1}{\sqrt{2nπ}}}=\sqrt{2π} \neq 1$ ${b_n}$ & ${a_n}$ have similar char. By p-series test* we can tell ${b_n}$ is divergent. Hence $a_{n}$ is divergernt
*p series test: For the series: $\sum_{n=1}^{\infty} \frac{1}{n^{p}}$ The $p$ -series converges if $p>1$ The $p$ -series diverges if $p \leq 1$ here p = 1/2