For which $x$ does the series $∑_{n=1}^∞(1+\frac{1}{2}+ \frac{1}{3}+⋯+\frac{1}{n})\, x^n$ converge? Determine for what value of $x$ the series converges
$$\sum_{n=1}^\infty \left(1+\frac{1}{2}+ \frac{1}{3}+⋯+\frac{1}{n}\right) x^n $$
Observe that 
$∑_{n=1}^∞(1+\frac{1}{2}+ \frac{1}{3}+⋯+\frac{1}{n}) x^n =∑_{n=1}^∞ (∑_{n=1}^∞\frac{1}{n}) x^n$
Can I consider $(\sum_{n=1}^\infty\frac{1}{n})$ as $a_n$ and use the ratio test here or I must use the Hadamard?

Let $H_n=1+\frac{1}{2}+\cdots+\frac{1}{n}$,
Then $\frac{H_{n+1}}{H_n}=1+\frac{1}{(n+1)H_n}$ tends to 1, Thus the ratio of convergence is $R=1$ (D'Alembert criterion).
Now note that the Cauchy product of 
$$
\sum_{n=0}^\infty x^n, \quad \sum_{n=1}^\infty  \frac{x^n}{n}
$$
is exactly 
$$
\sum_{n=1}^\infty H_n\, x^n
$$
Therefore,
$$
\forall x\in (-1,1), \sum_{n=0}^\infty H_n\, x^n=-\frac{ln(1-x)}{1-x}
$$
