[Math] Calculating infinite sums of power series.

calculuspower seriessequences-and-series

This past week I have been studying power series, first I studied how to determinate the intervals of convergence and I have no problem doing that (usually I just have to apply the root or ratio test of convergence). However, now I am asked to calculate the sums of: enter image description here

I know some results of infinite series, like the geometric or telescopic series, however this is not enough to calculate any of those infinite sums. Is there any general procedure to calculate this sums? Or any differentation/integration theorems of power series I could use?

Best Answer

Hints.

(a) $\dfrac{1}{n(2n-1)}=\dfrac{a}{n}+\dfrac{b}{2n-1}$.

(b) For $|z|< 1$, the derivative of $\sum_{n\geq 0} z^n=\frac{1}{1-z}$ is $\sum_{n\geq 1} nz^{n-1}=\frac{1}{(1-z)^2}$.

(c) $\dfrac{n^3}{n!}=\dfrac{an(n-1)(n-2)+bn(n-1)+c n}{n!}$.

(d) $\dfrac{1}{1+2+\dots +n}=\dfrac{2}{n(n+1)}=\dfrac{a}{n}+\dfrac{b}{n+1}$.

(e) $\dfrac{n^3+n+3}{n+1}=an^2+bn+c+\dfrac{d}{n+1}$.

Can you take it from here? Now you need to remember some basic power series. For example for (c), recall that for any real $x$, $\sum_{n\geq 0}x^n/n!=e^x$ and, by the above hint, we get $$\sum_{n\geq 0}\frac{n^3}{n!}x^n=ax^3\sum_{n\geq 3}\frac{x^{n-3}}{(n-3)!} +bx^2\sum_{n\geq 2}\frac{x^{n-2}}{(n-2)!}+cx\sum_{n\geq 1}\frac{x^{n-1}}{(n-1)!}=(ax^3+bx^2+cx)e^x. $$