Question

Find whether the following series are convergent or divergent:
$x^2{\left(\log 2\right)}^q+x^3{\left(\log 3\right)}^q+x^4{\left(\log 4\right)}^q+....$

Answer 1

Let $T_n$ be the general term of the series

$T_n=x^{n+1}{\left[\log (n+1)\right]}^q{T}_{n+1}=x^{n+2}{\left[\log (n+1)+1\right]}^q$

Applying ratio test

$\underset{n\to \infty }{lim}\frac{T_{n+1}}{T_n}=\underset{n\to \infty }{lim}\frac{x^{n+2}{\left[\log (n+1)+1\right]}^q}{x^{n+1}{\left[\log (n+1)\right]}^q}=\underset{n\to \infty }{lim}x{\left[\frac{(n+1)-\frac{{(n+1)}^2}{2}+\frac{{(n+1)}^3}{3}-...{(-1)}^{k+1}\frac{{(n+1)}^k}{k}}{n-\frac{n^2}{2}+\frac{n^3}{3}-\frac{n^4}{4}+....{(-1)}^{k+1}\frac{n^k}{k}}\right]}^q=\underset{n\to \infty }{lim}x\frac{{(n+1)}^{kq}}{n^{kq}}=\underset{n\to \infty }{lim}x{(1+\frac{1}{n})}^{kq}=x$

For $x>1$, the series is divergent

For $x<1$, the series is convergent

For $x=1$, the ratio test fails

We apply raabe's test

$\underset{n\to \infty }{lim}n[\frac{T_n}{T_{n+1}}-1]=\underset{n\to \infty }{lim}n[\frac{{\left[\log (n+1)\right]}^q}{{\left[\log (n+2)\right]}^q}-1]=\underset{n\to \infty }{lim}n\frac{{\left[\log (n+1)\right]}^q-{\left[\log (n+2)\right]}^q}{{\left[\log (n+2)\right]}^q}=\underset{n\to \infty }{lim}n\frac{[{(n-\frac{n^2}{2}+....{(-1)}^{k+1}\frac{n^k}{k})}^q-{\left(\right(n+1)-\frac{{(n+1)}^2}{2}+...{(-1)}^{k+1}\frac{{(n+1)}^k}{k})}^q]}{{\left[\right(n+1)-\frac{{(n+1)}^2}{2}+...{(-1)}^{k+1}\frac{{(n+1)}^k}{k}]}^q}=\underset{n\to \infty }{lim}\frac{n[n^{kq}-{(n+1)}^{kq}]}{{(n+1)}^{kq}}=\underset{n\to \infty }{lim}\frac{-\left(kq\right)n^{kq+1}}{n^{kq}}=-kq$

If $-kq>1\Rightarrow k>\frac{-1}{q}$, the series is convergent

If $-kq<1\Rightarrow k<\frac{-1}{q}$, the series is divergent

($k=$ positive integer)