Convergence of harmonic power series

Question

Find the interval of convergence for the series,

$$\sum_{n=0}^\infty \left(1+\frac{1}{2}+\ldots+\frac{1}{n}\right)\cdot x^n$$

I have no clue where to even being assessing this infinite series, however I can see that the harmonic series comes into play somehow. I am wondering if it's possible to rewrite the series as a nested, represented as the following,

$$\sum_{n=0}^\infty \left(\sum_{n=1}^\infty\left(\frac{1}{n}\right)\cdot x^n\right)$$

Would this be productive at all? I believe that it is necessary to use squeeze theorem to conclude that $\lim_{n\to\infty}\frac{|a_{n+1}|}{|a_n|} = |x|$, and it should also be shown that $a_n$ does not converge to $0$ if $|x|\geq1$. However, I have no clue to to approach either of these problems. Any help is greatly appreciated!

Answer 1

Hint

You correctly surmise that can use the ratio test to obtain the answer: \begin{align} \left|\frac{a_{n+1}}{a_n}\right|&=\left|\frac{\left(\sum_\limits{i=1}^{n+1}\frac{1}{i}\right)x^{n+1}}{\left(\sum_\limits{i=1}^n\frac{1}{i}\right)x^n}\right|\\ &=|x|\left(1+\frac{1}{(n+1)\sum_\limits{i=1}^n\frac{1}{i}}\right)\ . \end{align} What's the limit of this expression as $\ n\rightarrow\infty\ $?

Answer 2

Let $H_n=\sum_{k=1}^n\frac{1}{k}$ be the $n^{th}$ harmonic number. There is the famous Cauchy-Hadamard formula for the radius of convergence of a given power series, namely \begin{align} R=\frac{1}{\limsup_{n\to\infty}|H_n|^{1/n}} \end{align} In our case, note that the Harmonic numbers are roughly logarithmic, i.e once we get past the first few terms, we can find $c,C>0$ such that for all large enough $n$, $c\log n\leq H_n\leq C\log n$. This is essentially because $H_n=\sum_{k=1}^n\frac{1}{k}\sim\int_1^n\frac{1}{t}\,dt=\log n$. So, by using the fact that $\lim\limits_{n\to\infty}(c\log n)^{1/n}=1$ (since $c>0$) and likewise with $C\log n$, we see that $\lim\limits_{n\to\infty}H_n^{1/n}=1$. Hence, $R=1$.