Interval of convergence for $\sum_{n=1}^∞({1 \over 1}+{1 \over 2}+\cdots+{1 \over n})x^n$ What is the interval of convergence for $\sum_{n=1}^∞({1 \over 1}+{1 \over 2}+\cdots+{1 \over n})x^n$?
How do I calculate it? Sum of sum seems a bit problematic, and I'm not sure what rules apply for it. 
Thanks in advance. 
 A: The coefficient of $x^n$ is the $n$-th harmonic number $H_n$. One can show that there are positive constants $a$ and $b$ such that $a\ln n\lt H_n \lt b\ln n$. For much more detail than necessary, see the Wikipedia article on harmonic numbers. There will be no "endpoint" issue, since the coefficients go to $\infty$. 
More simply, one can use the Ratio Test directly on $H_n$. Since $H_n\to\infty$ as $n\to\infty$, one can show quickly that
$$\lim_{n\to\infty} \frac{H_{n+1}}{H_n}=1.$$
A: Let 
$$a_k=1\quad \forall k\geq 0$$
and
$$b_0=0\quad\text{and}\quad b_k=\frac{1}{k}\quad\forall k\geq 1$$
then the series $\displaystyle\sum_{k=0}^\infty a_k x^k=\frac{1}{1-x}$ and $\displaystyle\sum_{k=0}^\infty b_k x^k=-\log(1-x)$ are absolutely convergent and their radius of convergence is $R=1$ hence the series Cauchy product $\displaystyle\sum_{n=0}^\infty c_nx^n$ where
$$c_n=\sum_{k=0}^n a_{n-k}b_k=1+\frac{1}{2}+\cdots+\frac{1}{n}$$
 has also the radius of convergence $R=1$ and moreover we have
$$\sum_{k=1}^\infty \left(1+\frac{1}{2}+\cdots+\frac{1}{n}\right)x^n=\left(\sum_{k=0}^\infty a_k x^k \right)\times\left(\sum_{k=0}^\infty b_kx^k\right)=-\frac{\log(1-x)}{1-x}$$
