I am to show that the following series converges pointwise, then uniformly. I am aware that uniform convergence implies pointwise convergence, but I have to show the pointwise convergence first and separately.
$\sum_{n=1}^{\infty}log(1+\frac{x}{n^3}), \hspace{2cm} -1 < x <1$
I have started out by looking at the case of $x=0$, in which case the sum obviously converges to zero.
I am however, not entirely sure how to approach the series, as I can't seem to find another convergent series to use the comparison test on. I have tried doing a ratio test, but it ends up inconclusive, since
$lim_{n\to\infty}|\frac{log(1+\frac{x}{(n+1)^3})}{log(1+\frac{x}{n^3})}|=1$
As for the uniform convergence, I suppose I have to find a convergent majorantseries. But seeing as I struggle to determine a convergent series for a comparison test relating to the pointwise convergence, I am also at loss here.
Any input is appreciated.
For $0 \leqslant x < 1$ we have
$$\left|\log\left(1 + \frac{x}{n^3}\right)\right| = \log\left(1 + \frac{x}{n^3}\right) = \int_1^{1+x/n^3} \frac{dt}{t} \\\leqslant \frac{1+x/n^3-1}{1} = \frac{x}{n^3} \leqslant \frac{1}{n^3}$$
For $-1 < x < 0$ we have
$$\left|\log\left(1 + \frac{x}{n^3}\right)\right| = -\log\left(1 + \frac{x}{n^3}\right) = \int_{1+x/n^3}^{1} \frac{dt}{t} \\\leqslant \frac{1 - (1+x/n^3)}{1+x/n^3} = \frac{-x/n^3}{1+x/n^3} = \frac{1}{n^3/(-x) - 1},$$
and for all $n >1$ and $-1 < x < 0$,
$$\left|\log\left(1 + \frac{x}{n^3}\right)\right| < \frac{1}{n^3 - 1}$$
Thus, the $n$-th term is bounded above in absolute value by $1/(n^3-1)> 1/n^3$ for all $n> 1$ and for all $x \in (-1,1)$ and $\displaystyle\sum_{n > 1}\frac{1}{n^3-1}$ converges.
By the Weierstrass test the series is uniformly convergent.