13
$\begingroup$

What is the appropriate method to find the value of $1$ - $1\over 7$ + $1\over 13$ - ... upto infinite terms? (The denominators increase by 6 in consecutive terms)

I approximated it by integrating $\frac{1}{1+x^6}$ putting x=1...is there a better and nicer method ? :)But what should I take as limits of the integration?

$\endgroup$
14
  • 2
    $\begingroup$ Integrating that would be my first (may be only) idea. $\int_0^1x^6\,dx=1/7$, $\int_0^1x^{12}\,dx=1/13$... $\endgroup$ Sep 11, 2015 at 9:31
  • 1
    $\begingroup$ @SanchayanDutta That's correct $\endgroup$ Sep 11, 2015 at 9:40
  • 1
    $\begingroup$ I agree..hope to ask a question on that soon @OussamaBoussif :) $\endgroup$
    – user220382
    Sep 11, 2015 at 9:44
  • 1
    $\begingroup$ @SanchayanDutta You can think of it in even simpler terms than that: the series with terms $1, 0.4, 0.01, 0.004, 0.0002, \dots$ sums to $\sqrt{2}$ which is irrational. $\endgroup$ Sep 11, 2015 at 9:46
  • 4
    $\begingroup$ @SanchayanDutta … Only for finite sums. Infinite sums are inherently limiting processes, and limits have no inherent need to respect set membership. It's the same theorem as "strong inequalities may not pass to the limit". Equivalently, the rationals do not form a closed set in the metric space $\mathbb{R}$. $\endgroup$ Sep 11, 2015 at 9:49

4 Answers 4

7
$\begingroup$

For first, we have: $$\begin{eqnarray*}\sum_{n\geq 0}\frac{(-1)^n}{6n+1}&=&\sum_{n\geq 0}(-1)^n\int_{0}^{1}x^{6n}\,dx=\int_{0}^{1}\frac{dx}{1+x^6}\tag{1}\end{eqnarray*}$$ Now we may compute the last integral through partial fraction decomposition.

If $\xi_i$, $1\leq i\leq 6$, is a root of $1+x^6$, we have: $$\text{Res}\left(\frac{1}{x^6+1},x=\xi_i\right)=\frac{1}{6\xi_i^5}=-\frac{\xi_i}{6}\tag{2}$$ hence: $$ \int_{0}^{1}\frac{dx}{1+x^6}=-\frac{1}{6}\sum_{i=1}^{6}\int_{0}^{1}\frac{\xi_i}{x-\xi_i}\,dx=-\frac{1}{6}\sum_{i=1}^{6}\xi_i \log\left(1-\frac{1}{\xi_i}\right)\tag{3} $$ and: $$\begin{eqnarray*} \int_{0}^{1}\frac{dx}{1+x^6}&=&-\frac{1}{6}\sum_{j=0}^{5}e^{\frac{\pi i}{6}(2j+1)}\log\left(1-e^{-\frac{\pi i}{6}(2j+1)}\right)\\&=&-\frac{1}{6}\sum_{j=0}^{5}e^{\frac{\pi i}{6}(2j+1)}\left(-\frac{\pi i}{12}(2j+1)+\log\left(2i\sin\frac{\pi(2j+1)}{12}\right)\right)\\&=&\frac{\pi}{6}-\frac{1}{6}\sum_{j=0}^{5}e^{\frac{\pi i}{6}(2j+1)}\left(\frac{\pi}{2}i+\log\sin\frac{\pi(2j+1)}{12}\right)\\&=&\color{red}{\frac{\pi+\sqrt{3}\log(2+\sqrt{3})}{6}}.\tag{4}\end{eqnarray*}$$


Another possible approach is the following: we have $$ \sum_{n\geq 0}\frac{(-1)^n}{6n+1}=\sum_{n\geq 0}\left(\frac{1}{12n+1}-\frac{1}{12n+7}\right)=\frac{\psi\left(\frac{7}{12}\right)-\psi\left(\frac{1}{12}\right)}{12}\tag{5}$$ then the result follows from combining the reflection formula: $$ \psi(z)-\psi(1-z)=-\pi\cot(\pi z)\tag{6}$$ with the duplication formula: $$ \psi(z)+\psi\left(z+\frac{1}{2}\right)=-2\log 2+2\,\psi(2z)\tag{7}$$ and the triplication formula: $$ 3\,\psi(3z)=(3\log 3)z+\psi(z)+\psi\left(z+\frac{1}{3}\right)+\psi\left(z+\frac{2}{3}\right)\tag{8}$$ for the digamma function.

$\endgroup$
3
  • 2
    $\begingroup$ Undoubtedly elegant :)...but beyond my comprehension at present.. +1 $\endgroup$
    – user220382
    Sep 11, 2015 at 17:34
  • $\begingroup$ This looks like a great Answer, and I'd really like to understand it, but I couldn't follow several of your steps. $\endgroup$ Sep 11, 2015 at 21:38
  • $\begingroup$ It is just a partial fraction decomposition, carried on through the computation of some residues, followed by an explicit integration. $\endgroup$ Sep 11, 2015 at 21:42
6
$\begingroup$

I really like the approach that you would have liked to follow.

Here I give a yet another approach which might not be the most straightforward way. Taking a look here helps us a bit:

\begin{align} \sum_{n=0}^{\infty}\frac{(-1)^{n}}{1+6n}&=\frac16\sum_{n=0}^{\infty}\frac{(\color{blue}{-1})^{n}}{(n+\color{green}{\frac16})^\color{red}1}\\ &=\frac16\Phi(\color{blue}{-1},\color{red}1,\color{green}{\frac16})\\ &=\frac16\frac{1}{\Gamma(\color{red}1)}\int_0^{\infty}\frac{t^{\color{red}1-1}e^{-\color{green}{\frac16}t}}{1-(\color{blue}{-1})e^{-t}}dt\\ &=\frac16\int_0^{\infty}\frac{e^{-t/6}}{1+e^{-t}}dt\\ &=\int_0^{\infty}\frac{e^{-t}}{1+e^{-6t}}dt\\ &=\int_1^{\infty}\frac{t^4}{1+t^6}dt \end{align} where the last integral (which is manageable) results from $t\to \log t$.


Hint on solving the integral: \begin{align} \frac{t^4}{1+t^6}&=\frac{t^4}{(t^2+1)(t^4-t^2+1)}\\ &=\frac13\frac{1}{t^2+1}+\frac13\frac{2t^2-1}{t^4-t^2+1}\\ &=\frac13\frac{1}{t^2+1}+\frac13\frac{2t^2-1}{(t^2+1)^2-3t^2}\\ &=\frac13\frac{1}{t^2+1}+\frac13\frac{2t^2-1}{(t^2+1-\sqrt3t)(t^2+1+\sqrt3t)}\\ \end{align}

$\endgroup$
5
  • $\begingroup$ Seems like all sums over $\dfrac{z^n}{(n+\alpha)^\color{red}{1}}$ can be solved like this. Is the Lerch transcendent always manageable when the red number (power of the denominator) is 1? Mathematica gives nice functions in $z$ for all integer $\alpha$ I tried. $\endgroup$
    – Nikolaj-K
    Sep 11, 2015 at 11:45
  • $\begingroup$ It looks like this is just a detour on the integral $\int_{0}^{1}\frac{dx}{1+x^6}$, but the answer does not give a real hint about how to compute it. $\endgroup$ Sep 11, 2015 at 12:47
  • $\begingroup$ @NikolajK this is true, I agree. But I think one could think situations when the power is not one and still has a manageable integral on the desk ... $\endgroup$
    – Math-fun
    Sep 11, 2015 at 16:28
  • $\begingroup$ @JackD'Aurizio Many thanks! I agree with you that this is a detour. I simply enjoyed it :-) and wanted to share. About solving the integral: I will come back to this in the next days, I think one could do it a bit different from your (elegant) way. $\endgroup$
    – Math-fun
    Sep 11, 2015 at 16:43
  • $\begingroup$ @JackD'Aurizio its been some time since I last did something here ... now I updated the answer. $\endgroup$
    – Math-fun
    May 2, 2016 at 9:14
5
$\begingroup$

We can write $$\displaystyle S = 1-\frac{1}{7}+\frac{1}{13}-........\infty = \int_{0}^{1}\left(x^{0}-x^{6}+x^{12}-........\infty\right)dx$$

So $$\displaystyle S = \int_{0}^{1}\frac{1}{1+x^6}dx = \frac{1}{2}\int_{0}^{1}\frac{\left(1+x^4\right)+\left(1-x^4\right)}{1+x^6}dx$$

$$\displaystyle S = \frac{1}{2}\int_{0}^{1}\frac{1+x^4}{1+x^6}dx+\frac{1}{2}\int_{0}^{1}\frac{1-x^4}{1+x^6}dx$$

Now we will take $$\displaystyle I = \frac{1}{2}\int_{0}^{1}\frac{1+x^4}{1+x^6}dx$$ and $$\displaystyle J = \frac{1}{2}\int_{0}^{1}\frac{1-x^4}{1+x^6}dx$$

So first we will calculate value of $I$

So $$\displaystyle I = \frac{1}{2}\int_{0}^{1}\frac{1+x^4}{1+x^6}dx = \frac{1}{2}\int_{0}^{1}\frac{(x^2+1)^2-2x^2}{1+x^6}dx$$

So $$\displaystyle I = \frac{1}{2}\int_{0}^{1}\frac{(x^2+1)^2}{(1+x^2)\cdot (x^4-x^2+1)}dx - \int_{0}^{1}\frac{x^2}{1+(x^3)^2}dx$$

So $$\displaystyle I = \frac{1}{2}\int_{0}^{1}\frac{x^2+1}{x^4-x^2+1}-\int\frac{x^2}{1+(x^3)^2}dx$$

So $$\displaystyle I = \frac{1}{2}\int_{0}^{1} \frac{\left(1+\frac{1}{x^2}\right)}{\left(x-\frac{1}{x}\right)^2+1^2}- \int_{0}^{1}\frac{x^2}{1+(x^3)^2}dx$$

Now Let $$\displaystyle \left(x-\frac{1}{x}\right) = t \Leftrightarrow \left(1+\frac{1}{x^2}\right)dx = dt$$ and $x^3 = u\Leftrightarrow 3x^2dx = du\displaystyle \Leftrightarrow dx = \frac{1}{3}du$

So $$\displaystyle I = \frac{1}{2}\cdot \left[\tan^{-1}\left(x-\frac{1}{x}\right)\right]_{0}^{1} - \frac{1}{3}\cdot \left[\tan^{-1}\left(x^3\right)\right]_{0}^{1} = \frac{\pi}{4}-\frac{\pi}{12} = \frac{\pi}{6}$$

Similarly we will calculate for $$\displaystyle J = \frac{1}{2}\int_{0}^{1}\frac{1-x^4}{1+x^6}dx$$

So $$\displaystyle J = \frac{1}{2}\int_{0}^{1}\frac{(1-x^2)\cdot (1+x^2)}{(1+x^2)\cdot (x^4-x^2+1)}dx = -\frac{1}{2}\int_{0}^{1}\frac{x^2-1}{x^4-x^2+1}dx$$

$$\displaystyle J = -\frac{1}{2}\int_{0}^{1} \frac{\left(1-\frac{1}{x^2}\right)}{\left(x+\frac{1}{x}\right)^2+\left(\sqrt{3}\right)^2}dx$$

Now Now Let $$\displaystyle \left(x+\frac{1}{x}\right) = v \Leftrightarrow \left(1-\frac{1}{x^2}\right)dx = dv$$

So $$\displaystyle J = -\frac{1}{2}\cdot \frac{1}{2\sqrt 3}\cdot \left[\ln \left|\frac{\left(x+\frac{1}{x}\right)-\sqrt{3}}{\left(x+\frac{1}{x}\right)+\sqrt{3}}\right|\right]_{0}^{1} $$

$$\displaystyle J = -\frac{1}{2}\cdot \frac{1}{2\sqrt 3}\cdot \ln\left|\frac{x^2+1-\sqrt{3x}}{x^2+1+\sqrt{3}x}\right|_{0}^{1} = -\frac{2}{2}\cdot \frac{1}{2\sqrt 3}\cdot \ln(2-\sqrt{3}) = \frac{1}{2\sqrt{3}}\ln(2+\sqrt{3})$$

So $$\displaystyle \int_{0}^{1}\frac{1}{1+x^6}dx = \frac{\pi}{6}+\frac{1}{2\sqrt{3}}\ln(2+\sqrt{3}) = \color{red}{\frac{\pi+\sqrt{3}\ln(2+\sqrt{3})}{6}}$$

$\endgroup$
1
  • $\begingroup$ That was basically the integration part..isn't it ? Anyway I really appreciate the effort :)..thank you! $\endgroup$
    – user220382
    Sep 11, 2015 at 17:35
1
$\begingroup$

Define f(x) = x - $(x^7)\over 7$ +$ (x^{13})\over 13$ - ...

Note that f($0$)= $0$ and f($1$) is what you want to find.

You've already gathered that the derivative f'(x) = $1\over (1+x^6)$

$$ f(x)=\int_{?}^{x}\frac{1}{1+x'^{6}}dx' $$

Your lower limit needs to be $0$ so that f($0$) will come out to $0$. Your upper limit needs to be $1$ because f($0$) is what you're trying to find.

$$ \sum_{n=0}^{\infty}\frac{1}{1+6n}\left(-1\right)^{n}=\int_{0}^{1}\frac{1}{1+x'^{6}}dx' $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy