3

How can I compute the following series? \begin{equation} \sum_{n=1}^\infty\frac{n+10}{2n^2+5n-3}\left(-\frac{1}{3}\right)^n \end{equation}

I manipulated the term and got \begin{equation} \frac{n+10}{2n^2+5n-3}\left(-\frac{1}{3}\right)^n = \left(\frac{11}{14n-7}-\frac{1}{n+3}\right)\left(-\frac{1}{3}\right)^n\text{.} \end{equation} But I don't know what to do next.

zxcv
  • 1,445
  • 1
    @zxcv Hint: $\frac{d}{dx}(\frac{x^n}{n}) = x^{n-1}$ –  Jan 16 '21 at 05:13
  • @DeriveFoiler OP has already got it in the form of $x^n/(an+b)$ by separating the last line into two sigmas then it is easy after that to get into the required form as told by you. –  Jan 16 '21 at 05:15
  • 1
    @Forester I apologize; I was providing the next step. After that, the techniques I was referring to were either your hint or just plugging into $-\log(1-x)$ directly. I have deleted my comment because your explanation is a superset of mine. – Derive Foiler Jan 16 '21 at 05:20
  • @DeriveFoiler No need to apologize, sir. –  Jan 16 '21 at 05:30
  • The answer to the question in the title is that in general you can't – there's no closed form expression for, say, $\sum_1^{\infty}2^nn^{-2}$. – Gerry Myerson Jan 16 '21 at 05:34
  • @Forester So the key was differentiating (or integrating) the series. Thanks! – zxcv Jan 16 '21 at 05:37
  • @GerryMyerson Thanks. I thought it would be too specific for a title if I just wrote the series I want to compute. So I wrote it in a more general form, but I think it was too general. – zxcv Jan 16 '21 at 05:41
  • Nothing wrong with the title, it gave me a chance to write something that you might find interesting, or useful some day, or useful to someone else some day. – Gerry Myerson Jan 16 '21 at 06:06

2 Answers2

0

$$\frac{n+10}{2n^2+5n-3}=\frac{3}{2 n-1}-\frac{1}{n+3}$$

Consider the first sum $$S_1=\sum_{n=1}^\infty\frac{x^n}{2 n-1}=y\sum_{n=1}^\infty\frac{y^{2n-1}}{2 n-1}=y\tanh ^{-1}(y)=\sqrt{x} \tanh ^{-1}\left(\sqrt{x}\right)$$ With $x=-\frac 13$ we then have $$S_1=-\frac{\pi }{6 \sqrt{3}}$$

For the second sum $$S_2=\sum_{n=1}^\infty\frac{x^n}{ n+3}=\frac 1 {x^3}\sum_{n=1}^\infty\frac{x^{n+3}}{ n+3}=-\frac 1 {x^3}\left(\log(1-x)+x+\frac {x^2}2 +\frac {x^3}3\right)$$ With $x=-\frac 13$ we then have $$S_2=\frac{47}{6}-27\log \left(\frac{4}{3}\right)$$

For the total $$\frac{47}{6}-\frac{\pi }{2 \sqrt{3}}-27\log \left(\frac{4}{3}\right)$$

Claude Leibovici
  • 260,315
0

$$S=\sum_{n=1}^\infty\frac{n+10}{2n^2+5n-3}\left(-\frac{1}{3}\right)^n$$ $$S=\sum_{n=1}^{\infty} \left(\frac{3}{2n-1}-\frac{1}{n+3}\right)(-3^{-1})^{n}=S_1-S_2$$ $$S_2=-27\sum_{n=1}^{\infty} \frac{(-3^{-1})^{n+3}}{n+3}$$ Use $\ln(1+z)=\sum_{k=1}^{\infty} (-1)^{k-1} \frac{z^k}{k}$ $$=-27\sum_{m=4}^{\infty}\frac{(-1)^m(3^{-1})^m}{m}=\frac{-47}{6}+27\ln(4/3)$$ Next use $\tan^{-1} x=\sum_{k=1}^{\infty} (-1)^{k-1}\frac{z^{2k-1}}{2k-1}$ $$S_1=\sqrt{3}\sum_{n=1}^{\infty} (-1)^n \frac{(1/\sqrt{3})^{2n-1}}{2n-1}=-\sqrt{3} \tan^{-1}(1/\sqrt{3})=\frac{-\pi}{2 \sqrt{3}}$$ Finally $$S=\frac{-\pi}{2 \sqrt{3}}+\frac{47}{6}-27\ln(4/3)$$

Z Ahmed
  • 43,235