Determine for what values of $x$ the given series converges The given series is 

$\sum_{n=1}^∞
(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+...+\frac{1}{n}
)x^{n} $.

I tried it by using Cauchy Root Test as follows-
Let 
$y=\lim_{n\to\infty}(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+...+\frac{1}{n} )^{1/n}$,
then by taking logarithm both sides,we get
$\log(y)=\lim_{n\to\infty}\frac{1}{n}\log(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+...+\frac{1}{n} )$
Since, $\log(0)=-\infty$. So, $\log(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+...+\frac{1}{n} )=-\infty$.
Applying,L'Hôpital's rule,we get 
$\log(y)=\lim_{n\to\infty}-\frac{(\frac{1}{n^2}+\frac{1}{(n-1)^2}+\frac{1}{(n-2)^2}+...+...+... )}{(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+...+\frac{1}{n} )}$.(Please Check this step!!)
Since,$\sum_{k=1}^\infty\frac{1}{k^2}=\lim_{n\to\infty}\sum_{k=1}^n\frac{1}{k^2}=\frac{\pi^2}{6}$ & $\sum_{k=1}^\infty\frac{1}{k}=\lim_{n\to\infty}\sum_{k=1}^n\frac{1}{k}=\infty$.So,$\log(y)=0\implies y=1$.
Now let $a_n=(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+...+\frac{1}{n} )x^{n} $. Then,$$(a_n)^{1/n}=\lim_{n\to\infty}(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+...+\frac{1}{n} )^{1/n}x$$
$\implies \lim_{n\to\infty}\vert (a_n)^{1/n}\vert=1.x$.By ,Cauchy root test the given series converges if $\lim_{n\to\infty}(a_n)^{1/n}<1$.Hence, the given series converges if $\vert x\vert<1$.
I  NEED TO KNOW WHETHER MY SOLUTION IS CORRECT OR NOT?
 A: There are two serious problems. The first one is your assertion that
$$
\log\Bigl(1+\frac12+\dots+\frac1n\Bigr)=-\infty.
$$
It is well known (harmonic series) that
$$
\lim_{n\to\infty}\Bigl(1+\frac12+\dots+\frac1n\Bigr)=+\infty,
$$
so that
$$
\lim_{n\to\infty}\log\Bigl(1+\frac12+\dots+\frac1n\Bigr)=+\infty.
$$
The second one is about the use of L'Hôpital's rule. It is incorrect by two reasons: 1) you have a sequence, not a function and 2) the index of summation is also a variable.
There are several ways to solve the problem. For example, from the inequality
$$
1\le1+\frac12+\dots+\frac1n\le n
$$
it follows that
$$
\lim_{n\to\infty}\Bigl(1+\frac12+\dots+\frac1n\Bigr)^{1/n}=1.
$$
Or you can use the well known fact that
$$
1+\frac12+\dots+\frac1n\sim\log n.
$$
A: Since both $\frac{1}{1-x}=\sum_{n\geq 0}x^n$ and $-\log(1-x)=\sum_{n\geq 1}\frac{x^n}{n}$ have radius of convergence equal to one, the same holds for their (Cauchy) product
$$ \sum_{n\geq 1} H_n x^n = -\frac{\log(1-x)}{1-x}. $$
In particular the LHS is convergent for any $x\in(-1,1)$ and clearly is not convergent for any $x\in\mathbb{R}\setminus(-1,1)$ since in such a case the main term is not $o(1)$.
