What interval (of x) does this power series converge for? $$\sum_{n=1}^\infty \left(1+\frac{1}{2}+\dots+\frac{1}{n}\right)x^n$$
I think we can write it in this form also:
$\sum_{n=1}^\infty \left(\sum_{k=1}^{n}\frac{1}{k}\right)x^n$
I tried the ratio test:
$$\lim_{n\to\infty} \left|\frac{\left(\sum_{k=1}^{n+1}\frac{1}{k}\right)x^{n+1}}{\left(\sum_{k=1}^{n}\frac{1}{k}\right)x^{n}}\right|$$
Here, I don't know what to do with those summations, because they're both infinity.
 A: If $x\in [1,\infty),$ then $\sum_{n=1}^\infty \left(1+\frac{1}{2}+\dots+\frac{1}{n}\right)x^n\ge \sum_{n=1}^\infty x^n$ so the series diverges there.
So suppose $0\le x<1.$ Then, since $2+\frac{1}{3}+\dots+\frac{1}{n+1}\le\int_1^{n+1}\frac{1}{t}dt,$ we have
$\sum_{n=1}^\infty \left(1+\frac{1}{2}+\dots+\frac{1}{n}\right)x^n\le x^n-\frac{x^{n+1}}{n+1}+x^n\ln (n+1)$
so we conclude that the series converges (the last term converges, for example, by L'Hospital's rule.)
Now we are done, because the series has an interval of convergence symmetric about $x=0,$ so the foregoing implies that it converges in $(-1,1).$
A: For any integer $n\ge1,$
$$H_n:=\sum_{k=1}^n\frac1k\ge1$$ so, as $n\to+\infty,$
$$\frac{H_{n+1}}{H_n}=1+\frac1{(n+1)H_n}\to1$$
and
$$\frac{H_{n+1}|x|^{n+1}}{H_n|x|^n}\to|x|$$
hence the radius of convergence of this power series is equal to $1.$
Therefore, the series is absolutely convergent when $|x|<1,$ and trivially divergent (i.e. its general term does not tend to $0$) when $|x|>1.$
When $x=\pm1,$ it is also trivially divergent.
See the sum of this power series.
