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I have a trouble with calculating the sum of this series:

$$2+\sum_{n=1}^{\infty}\frac{1-n}{9n^3-n}$$

I tried to split it into three separate series like this: $$2+\sum_{n=1}^{\infty}\frac{1-n}{9n^3-n} =2+\sum_{n=1}^{\infty}\frac{2}{3n+1}+\sum_{n=1}^{\infty}\frac{1}{3n-1}-\sum_{n=1}^{\infty}\frac{1}{n} $$ but I'm not able to continue, can you give me some tips?

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    $\begingroup$ According to WolframAlpha the result is $\frac{\pi}{6 \sqrt 3} + \frac{3 \log(3)}{2}$ so this will take some tricks to work ... $\endgroup$ – Andreas May 23 at 19:57
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Let's rewrite $\frac{1}{n}$ as $\frac{3}{3n}$, so your sum is $S+2$ with$$S:=\sum_{n\ge 1}\left(\frac{1}{3n-1}-\frac{3}{3n}+\frac{2}{3n+1}\right)\\=\sum_{n\ge 1}\int_0^1 x^{3n-2}\left(1-3x+2x^2\right)\mathrm dx=\int_0^1\frac{x-3x^2+2x^3}{1-x^3}\mathrm dx.$$Thus $$S+2=\int_0^1\frac{2+3x}{1+x+x^2}\mathrm dx=\frac32\int_0^1\frac{1+2x}{1+x+x^2}\mathrm dx+\frac12\int_0^1\frac{\mathrm dx}{1+x+x^2}.$$The rest I'll leave to you.

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    $\begingroup$ Could you explain the first step? Why there is x^(3n-2)? Thank you. $\endgroup$ – aaa May 23 at 20:17
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    $\begingroup$ @aaa Because $\int_0^1 x^{k-1}dx=\frac{1}{k}$. $\endgroup$ – J.G. May 23 at 20:20
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    $\begingroup$ nicely done and +1 $\endgroup$ – logo May 23 at 22:14
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For the direct evaluation of the limit, you have received the good solution from J.G.

You could also consider the partial sum using, as you did, partial fraction decomposition $$\frac{1-n}{9n^3-n}=\frac{1}{3 n-1}+\frac{2}{3 n+1}-\frac{1}{n}$$ which makes $$S_p=\sum_{n=1}^{p}\frac{1-n}{9n^3-n}=\frac{1}{3} \left(\psi \left(p+\frac{2}{3}\right)-\psi \left(\frac{2}{3}\right)\right)+\frac{2}{3} \left(\psi \left(p+\frac{4}{3}\right)-\psi \left(\frac{4}{3}\right)\right)-H_p$$ where appear the digamma function and harmonic number.

Now, using asymptotics and Taylor series you should arrive to $$S_p=-\left(\gamma +\frac{1}{3}\psi\left(\frac{2}{3}\right)+\frac{2}{3} \psi \left(\frac{4}{3}\right)\right)+\frac{1}{9 p}-\frac{1}{9 p^2}+O\left(\frac{1}{p^3}\right)$$ which shows the limit and how it is approached.

Now, the particular values $$\psi\left(\frac{2}{3}\right)=-\gamma +\frac{\pi }{2 \sqrt{3}}-\frac{3 \log (3)}{2}$$ $$\psi\left(\frac{4}{3}\right)=3-\gamma -\frac{\pi }{2 \sqrt{3}}-\frac{3 \log (3)}{2}$$ make $$S_p=\left(-2+\frac{\pi }{6 \sqrt{3}}+\frac{3 \log (3)}{2}\right)+\frac{1}{9 p}-\frac{1}{9 p^2}+O\left(\frac{1}{p^3}\right)$$

For $p=10$, the exact value is $-\frac{477820712081}{12033629407800}\approx -0.039707$ while the above truncated expansion gives $-\frac{199}{100}+\frac{\pi }{6 \sqrt{3}}+\frac{3 \log (3)}{2}\approx -0.039782$.

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