Question

Find whether the following series is convergent or divergent:
$1+\frac{1}{2^2}+\frac{2^2}{3^3}+\frac{3^3}{4^4}+\frac{4^4}{5^5}+......$

Answer 1

Given $1+\frac{1}{2^2}+\frac{2^2}{3^3}+\frac{3^3}{4^4}+\frac{4^4}{5^5}+...$

Let $a_n=$ general term of series

$a_n=1+\frac{n^n}{(n+1{)}^{n+1}}{a}_{n+1}=1+\frac{(n+1{)}^{n+1}}{(n+2{)}^{n+2}}$

Applying ratio test

$\underset{n\to \infty }{lim}\frac{a_{n+1}}{a_n}=\underset{n\to \infty }{lim}\left[\frac{{(n+2)}^{n+2}+{(n+1)}^{n+1}}{{(n+2)}^{n+2}}\frac{{(n+1)}^{n+1}}{{(n+1)}^{n+1}+n^n}\right]=\underset{n\to \infty }{lim}\frac{{\left(n\right)}^{n+2}{(1+2/n)}^{n+2}+{\left(n\right)}^{n+1}{(1+1/n)}^{n+1}}{{\left(n\right)}^{n+2}{(1+2/n)}^{n+2}}\times \frac{{\left(n\right)}^{n+1}{(1+1/n)}^{n+1}}{{\left(n\right)}^{n+1}{(1+1/n)}^{n+1}+n^n}=\underset{n\to \infty }{lim}\frac{[\frac{n^{n+2}}{n^{n+2}}{(1+2/n)}^{n+2}+\frac{n^{n+1}}{n^{n+2}}{(1+1/n)}^{n+1}]\frac{n^{n+1}}{n^{n+1}}{(1+1/n)}^{n+1}}{{(1+2/n)}^{n+2}({(1+1/n)}^{n+1}+\frac{n^n}{n^{n+1}})}=\underset{n\to \infty }{lim}\frac{[{(1+2/n)}^{n+2}+\frac{1}{n}{(1+1/n)}^{n+1}]{(1+1/n)}^{n+1}}{{(1+2/n)}^{n+2}({(1+1/n)}^{n+1}+\frac{1}{n})}$

$=\underset{n\to \infty }{lim}\frac{(e^{\frac{2}{n}(n+2)}+0)e^{\frac{1}{n}(n+1)}}{e^{\frac{2}{n}(n+2)}{e}^{\frac{1}{n}(n+1)}}=1$ (inconclusive)

Series may be convergent or divergent

Ratio test fails

Now applying Raabel's test

$=\underset{n\to \infty }{lim}n(\frac{a_n}{a_{n+1}}-1)=R=\underset{n\to \infty }{lim}n[\frac{{(n+1)}^{n+1}+n^n}{{(n+1)}^{n+1}}\times \frac{{(n+2)}^{n+2}}{{(n+2)}^{n+2}+{(n+1)}^{n+1}}-1]=\underset{n\to \infty }{lim}n\left[\frac{n^n{(n+2)}^{n+2}-{(n+1)}^{2n+2}}{{(n+1)}^{n+1}({(n+2)}^{n+2}+{(n+1)}^{n+1})}\right]=\underset{n\to \infty }{lim}n\left[\frac{n^n{(n+2)}^{n+2}{(1+\frac{2}{n})}^{n+2}+n^{2n+2}{(1+\frac{1}{n})}^{2n+2}}{n^{n+1}{(1+\frac{1}{n})}^{n+1}{n}^{n+2}{(1+\frac{2}{n})}^{n+2}+n^{n+1}{(1+\frac{1}{n})}^{n+1}}\right]=\underset{n\to \infty }{lim}n\left[\frac{n^{2n+2}}{n^{2n+2}\cdot n}\frac{{(1+\frac{2}{n})}^{n+2}+{(1+\frac{1}{n})}^{2n+2}}{{(1+\frac{1}{n})}^{n+1}[{(1+\frac{2}{n})}^{n+2}+{(1+\frac{1}{n})}^{n+1}]}\right]$

$=\frac{e^2+e^2}{e\left(e^2\right)}=\frac{2e^2}{e^3}=\frac{2}{e}<1$ Divergent