I'm asked to find the interval of convergence of this series: $$\sum_{n=0}^\infty \frac{{x}^{3n}}{n+1}$$
and then the sum of the series.
Attempted Solution:
Clearly the series converges when $\vert x \vert<1$.
I put $x^3=y$ so that now we have:
$$\sum_{n=0}^\infty \frac{{y}^{n}}{n+1}$$
knowing that $\frac{y^n}{n+1} = \frac{1}{y}* \int y^ndy$ we can rewrite the series as: $$\sum_{n=0}^\infty \frac{{y}^{n}}{n+1} = \frac{1}{y}\sum_{n=0}^\infty\int y^ndy= \frac{1}{y}\int\sum_{n=0}^\infty y^n = \frac{1}{y}\int_{0}^{\infty}\frac{1}{1-y} =\frac{-1}{y}*ln(1-y) $$
In conclusion $$\sum_{n=0}^\infty \frac{{x}^{3n}}{n+1}=\frac{-1}{x^3}*ln(1-x^3) $$
Does this make any sense?
You have the right idea, but execution is a little sloppy. Start with $$\frac{y^n}{n+1}=\frac1y\int_0^yt^ndt$$ Then $$\sum_{n=0}^{\infty}\frac{y^n}{n+1}=\frac1y\sum_{n=0}^{\infty}\int_0^yt^ndt=\frac1y\int_0^y\sum_{k=0}^{\infty}t^ndt=\frac1y\int_0^y\frac{dt}{1-t}=-\frac1y\ln(1-y)$$ Where we have relied on uniform convergence of the sum for $0<y<1$.
See the difference? You didn't know what to take as the upper limit for the integral because $y$ was already the variable of integration so you chose $\infty$ and actually created a divergent integral, which could invalidate your proof.