Question

Find whether the following series are convergent or divergent:
$\frac{a+x}{1}+\frac{{(a+2x)}^2}{|\underset{–}{2}}+\frac{{(a+3x)}^3}{|\underset{–}{3}}+...$

Answer 1

Let $T_n$ be the general term of the given series

$T_n=\frac{{(a+nx)}^n}{n!}{T}_{n+1}=\frac{{(a+(n+1\left)x\right)}^{n+1}}{(n+1)!}$

Applying ratio test

$\underset{n\to \infty }{lim}\frac{T_{n+1}}{T_n}=\underset{n\to \infty }{lim}\frac{{(a+(n+1\left)x\right)}^{n+1}}{{(a+nx)}^n}\frac{n!}{(n+1)!}\underset{n\to \infty }{lim}\frac{n^{n+1}}{n^n}\frac{{(x+(x+a/n\left)\right)}^{n+1}}{{(x+a/n)}^{n}n(1+\frac{1}{n})}=\frac{x^{n+1}}{x^n}=x$

If $x>1$, the series is divergent

If $x<1$, the series if convergent

For $x=1$, the ratio test fails

We apply Raabe's test

$\underset{n\to \infty }{lim}n[\frac{T_n}{T_{n+1}}-1]=\underset{n\to \infty }{lim}n[\frac{{(a+n)}^n(n+1)}{{(a+(n+1\left)\right)}^{n+1}}-1]=\underset{n\to \infty }{lim}n\left[\frac{{(a+n)}^n(n+1)-{(a+(n+1\left)\right)}^{n+1}}{{(a+(n+1\left)\right)}^{n+1}}\right]=\underset{n\to \infty }{lim}n\frac{\left[n^{n+1}{[{(\frac{a}{n}+1)}^n(1+\frac{1}{n})-[\frac{a}{n}+(1+\frac{1}{n})]]}^{n+1}\right]}{\left[n^{n+1}{[\frac{a}{n}+(1+\frac{1}{n})]}^{n+1}\right]}=\underset{n\to \infty }{lim}n\frac{(e^a-e^{a+1})}{e^{a+1}}=\infty >1$

The series is convergent for $x=1$