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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?

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    $\begingroup$ For a, b, d, e, start with the well known geometric series $1+x+x^{2}+...=\frac {1}{1-x}$ (you should consider an appropriate expression that will be substituted to x.) then use differentiation or integration of power series, for (c), use the Maclaurin series of $e^{x} $. $\endgroup$ – 민찬홍 Nov 18 '17 at 16:03
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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 $$\sum_{n\geq 0}\frac{n^3}{n!}x^n=ax^3\sum_{n\geq 3}\frac{x^{n-3}}{(n-3)!} +bx^2\sum_{n\geq 2}\frac{x^{n-2}}{(n-2)!}+cx\sum_{n\geq 1}\frac{x^{n-1}}{(n-1)!}=(ax^3+bx^2+cx)e^x. $$

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\begin{align} \sum_{n=1}^\infty \frac{n(x+3)^n}{2^n} & = (x+3)\sum_{n=1}^\infty \frac{n(x+3)^{n-1}}{2^n} = (x+3)\sum_{n=1}^\infty \frac d {dx} \, \frac{(x+3)^n}{2^n} \\[10pt] & = (x+3) \frac d {dx} \sum_{n=1}^\infty \frac{(x+3)^n}{2^n} \qquad \text{See the comment on this step below.} \\[10pt] & = (x+3) \frac d {dx} \, \frac{\text{first term}}{1 - \text{common ratio}} = \text{etc.} \end{align} Is the sum of the derivatives equal to the derivative of the sum? In first-semester calculus you see it proved that that is true when there are only finitely many terms. It doesn't entirely generally work with inifinitely many terms, but it does with power series in the interior of the interval of convergence.

\begin{align} n^3 x^n & = x^3 \cdot\Big( n(n-1)(n-2)x^{n-3}\Big) + 3x^2 \cdot \Big( n(n-1) x^{n-2} \Big) + 3x \cdot\Big( nx^{n-1}\Big) \\[10pt] & = x^3 \frac {d^3} {dx^3} x^n + 3x^2 \frac{d^2}{dx^2} x^n + 3x \frac d {dx} x^n \\[12pt] \text{So } \sum_{n=0}^\infty \frac{n^3 x^n}{n!} & = x^3 \frac{d^3}{dx^3} \sum_{n=0}^\infty \frac{x^n}{n!} + 3x^2 \frac{d^2}{dx^2} \sum_{n=0}^\infty \frac{x^n}{n!} + 3x \frac d {dx} \sum_{n=0}^\infty \frac{x^n} {n!} \end{align} And presumably you know how to sum the series in the last line above.

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