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Can anyone help me to find what is the value of $1 -\frac{1}{4} + \frac{1}{7} - \frac{1}{10} + \frac{1}{13} - \frac{1}{16} + \frac{1}{19} + ... $ when it tends to infinity

The first i wanna find the pattern but it seems do not have any unique pattern can anyone help me?

Olivier Oloa
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Deddy
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3 Answers3

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Hint. Observe that the general term of your series is $$\frac{(-1)^n}{3n+1}. $$ Then you may write $$ \begin{align} \sum_{n=0}^{\infty}\frac{(-1)^n}{3n+1}&=\sum_{n=0}^{\infty}(-1)^n\int_0^1 x^{3n} dx\\\\ &=\int_0^1 \sum_{n=0}^{\infty}(-1)^nx^{3n} dx\\\\ &=\int_0^1 \frac{1}{1+x^3} dx\\\\ &=\int_0^1 \left(\frac{1}{3 (1+x)}+\frac{2-x}{3 \left(1-x+x^2\right)}\right) dx\\\\ &=\frac{1}{3}\ln 2+\frac{\pi\sqrt{3}}{9} \end{align} $$

Some details. $$ \begin{align} \int_0^1 \frac{1}{1+x^3} dx&=\frac{1}{3}\int_0^1\!\frac{dx}{(1+x)}+\frac{1}{3}\int_0^1\!\frac{2-x}{(x-1/2)^2+3/4} dx\\\\ &=\frac{1}{3}\ln 2+\frac{1}{3}\int_{-1/2}^{1/2}\!\frac{3/2-u}{u^2+3/4} du \quad (u=x-1/2, \,dx=du)\\\\ &=\frac{1}{3}\ln 2+\frac{1}{3}\int_{-1/2}^{1/2}\!\frac{3/2}{u^2+3/4} du\\\\ &=\frac{1}{3}\ln 2+\int_{0}^{1/2}\!\frac{1}{u^2+3/4} du\\\\ &=\frac{1}{3}\ln 2+\left.\frac{2}{\sqrt{3}}\arctan \left( \frac{2}{\sqrt{3}}u\right)\right|_{0}^{1/2} \\\\ &=\frac{1}{3}\ln 2+\frac{\pi\sqrt{3}}{9} \end{align} $$ where we have used $\displaystyle \arctan \!\left(\! \frac{1}{\sqrt{3}}\!\right)\!=\frac{\pi}{6}$.

Olivier Oloa
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  • You are using the Dominated Convergence Theorem at the second row. Great solution Olivier – Paolo Leonetti Feb 15 '15 at 09:21
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    Thanks @PaoloLeonetti Another way is to just "control" the remainder here: $$ \int_0^1 \frac{x^{3N}}{1+x^3} dx\leq \int_0^1 x^{3N} dx=\frac{1}{3N+1} \to 0$$ as $N \to +\infty$. – Olivier Oloa Feb 15 '15 at 09:24
  • @OlivierOloa do you have any solution without integral? Such using telescoping only maybe – Deddy Feb 15 '15 at 16:37
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    @Deddy If you know the digamma function (http://mathworld.wolfram.com/DigammaFunction.html), you can use it here, but this would be using integral in disguise... I think that if another solution is given, it will use integral, at least in disguise. Observe that these integrals are elementary ones. Telescoping sums can't hold here since the series sum is not rational. Thanks. – Olivier Oloa Feb 15 '15 at 16:50
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the series equal to $$\sum_{n=0}^{\infty}\frac{1}{6n+1}-\frac{1}{6n+4}$$ the first term $$\frac{1}{1-x^6}=\sum_{n=0}^{\infty} x^{6n}$$ $$\int_0^1\frac{1}{1-x^6}dx=\int_0^1\sum_{n=0}^{\infty}x^{6n}dx$$ $$\int_0^1\frac{1}{1-x^6}dx=\sum_{n=0}^{\infty}\frac{1}{6n+1}$$

the second term

$$\frac{1}{1-x^6}=\sum_{n=0}^{\infty}x^{6n}$$ mutiply by $x^3$

$$\frac{x^3}{1-x^6}=\sum_{n=0}^{\infty}x^{6n+3}$$ $$\int_0^1\frac{x^3}{1-x^6}dx=\int_0^1\sum_{n=0}^{\infty}x^{6n+3}dx$$ $$\int_0^1\frac{x^3}{1-x^6}dx=\sum_{n=0}^{\infty}\frac{1}{6n+4}$$

hence $$\sum_{n=0}^{\infty}\frac{1}{6n+1}-\frac{1}{6n+4}=\int_0^1\frac{1}{1-x^6}dx-\int_0^1\frac{x^3}{1-x^6}dx=\int_0^1\frac{1}{1+x^3}dx$$ $$=\frac{1}{3}\log 2+\frac{\pi\sqrt{3}}{9}$$

  • Is there any solution without integral? – Deddy Feb 15 '15 at 11:23
  • can i have the proof of this one? $\int_0^1\frac{1}{1+x^3}dx = =\frac{1}{3}\log 2+\frac{\pi\sqrt{3}}{9}$ – Deddy Feb 15 '15 at 14:23
  • The trouble with this solution is, the series $\sum_{n=0}^{\infty}\frac{1}{6n+1}$ and $\sum_{n=0}^{\infty}\frac{1}{6n+4}$ are divergent. So you are evaluating $\infty-\infty$, like a physicist or something. – TonyK Feb 16 '15 at 11:27
  • @TonyK: isn't math interesting enough? Do we really need rudeness wherever we go? – Gottfried Helms Feb 17 '15 at 00:18
  • But @Gottfried, it's what physicists do. They call it renormalisation. – TonyK Feb 17 '15 at 06:49
  • @TonyK: so it seems, that -at least- someone needs it. It is as it had to be expected. MSE- welcome sci.math... – Gottfried Helms Feb 17 '15 at 07:21
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Look at the final answer. The final answer tells you this series cannot be calculated simply. Here I propose a solution although it is not a good solution:

$$\Sigma_{n=1}^\infty \frac1{(n+a)(n+b)}=\frac{H_a-H_b}{a-b}$$

Where $H_x$ is harmonic number of $x$.

(It can be replaced with digamma function too. Although $\psi_0(x)\neq H_x$, the result is remained the same.)

Computing $H_x$ is not easy too. It is as hard as calculation of integrals. But if you have its table, you can calculate the series.

In this case, if you know $H_{\frac16}$ and $H_{\frac23}$ you can solve this series (Odd condition!).

$$\Sigma_{n=0}^\infty \frac{(-1)^n}{(3n+1)}=\Sigma_{n=0}^\infty (\frac1{(6n+1)}-\frac1{(6n+4)})=\Sigma_{n=0}^\infty \frac3{(6n+1)(6n+4)}=1-\frac14+\frac3{36}\Sigma_{n=1}^\infty \frac1{(n+\frac16)(6n+\frac46)}=1-\frac14+\frac1{12}\times\frac{H_{\frac46}-H_{\frac16}}{\frac46-\frac16}=\frac34+\frac16(H_{\frac23}-H_{\frac16})$$

I need to know these two harmonic numbers (the big disadvantage of this method):

$$H_{\frac23}=\frac32+\frac\pi{2 \sqrt3}-\frac{3 \log 3}2$$

$$H_{\frac16}=6-\frac{\sqrt3 \pi}2-\frac{3 \log 3}2-\log4$$

$$H_{\frac23}-H_{\frac16}=-\frac92+\frac{2\pi \sqrt3}3+2\log 2$$

$$\Sigma_{n=0}^\infty \frac{(-1)^n}{(3n+1)}=\frac34+\frac16 (-\frac92+\frac{2\pi \sqrt3}3+2\log 2)$$

$$\therefore ~~~\Sigma_{n=0}^\infty \frac{(-1)^n}{(3n+1)}=\frac{\pi \sqrt3}9+\frac{\log2}3$$

Arashium
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  • can i know where can i find the list of the harmonic number? – Deddy Feb 16 '15 at 07:11
  • For numbers higher than 1 there are some propose numerical methods and there are some numerical tables too. However, you need tables for simple fraction of small numbers and in symbolic way. Although they can be found in wolframalpha, a neat table of collection of them is something that I have not found on web. Maybe we need a pioneer to do that! :D – Arashium Feb 16 '15 at 11:03