What is the appropriate method to find the value of $1$ - $1\over 7$ + $1\over 13$ - ... upto infinite terms?

sequences-and-series

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?

Solution 1:

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.

Solution 2:

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}

Solution 3:

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