Calculus Problem by LOGA 5

Since $\frac{a_{n+1}}{a_n} \; = \; 1 + \frac{1}{na_n}$ we see that $\left| \frac{a_{n+1}}{a_n} - 1\right| \; = \; \frac{1}{na_n} \; \le \; n^{-1}$ and hence $\lim_{n \to \infty}\frac{a_{n+1}}{a_n} \; = \; 1$ Thus we deduce that the radius of convergence of this power series is $1$ . Since $a_n \to \infty$ as $n \to \infty$ , neither $\sum_{n \ge 0}a_n$ nor $\sum_{n\ge 0}(-1)^na_n$ converge. and so this power series converges for real $-1 < x < 1$ .

$f(x)=1(x+x^2+x^3+\cdots) + \frac{1}{2}(x^2+x^3+x^4+\cdots) + \frac{1}{3}(x^3+x^4+x^5+\cdots)+\cdots\\=(x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+\cdots)(1+x+x^2+x^3+\cdots)\\=-\log(1-x)\cdot \frac{1}{1-x}$ where both the convergence ranges are $(-1,1)$