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Finding value of $$1+\frac{1}{5}-\frac{1}{7}-\frac{1}{11}+\frac{1}{13}+\frac{1}{17}-\frac{1}{19}-\frac{1}{23}+\cdots$$

Try: I have solved in using Integration

But i am trying to solver it without using integration

Witting above series as $$\sum^{\infty}_{n=0}(-1)^n\bigg[\frac{1}{6n+1}+\frac{1}{6n+5}\bigg]$$

Now i am trying to solve it using euler reflection formula

Could not find any clue

could some help me to solve it

DXT
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    Would writing it as $1+\sum^{\infty}_{n=1} (-1)^{n}\bigg[\frac{1}{6n+1}-\frac{1}{6n-1}\bigg]$ help? – For the love of maths Jan 28 '19 at 06:59
  • I would try to perform multisection on the Maclaurin series of $\ln(1-x)$, to obtain closed form of the series $$\sum \frac{x^{6n+1}}{6n+1}$$, and then plug in a suitable value of $x$. – Szeto Jan 28 '19 at 07:51
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    Is there anything I could improve which prevents you right now from accepting my answer? – mrtaurho May 15 '19 at 09:29

1 Answers1

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$$1+\frac15-\frac17-\frac1{11}+\frac1{13}+\frac1{17}-\frac1{19}-\frac1{23}+\cdots=\frac{\pi}3$$

Utilizing the observation done by Mohammad Zuhair Khan we can explicitly write down the series as

$$1+\sum_{n=1}^\infty(-1)^n\left[\frac1{6n+1}-\frac1{6n-1}\right]=1-2\sum_{n=1}^\infty \frac{(-1)^n}{36n^2-1}$$

The latter form can be rewritten such that we can apply a sum identity of the cosecant function. To be precise we will use the formula

$$\pi\csc(\pi z)=z\sum_{n=-\infty}^\infty \frac{(-1)^n }{z^2-n^2}=\frac1z-2z\sum_{n=1}^\infty \frac{(-1)^n}{n^2-z^2}$$

Thus, lets rewrite the given series in the following way

$$\begin{align} 1-2\sum_{n=1}^\infty\frac{(-1)^n}{36n^2-1}&=1+\frac16\left[-2\frac16\sum_{n=1}^\infty \frac{(-1)^n}{n^2-\left(\frac16\right)^2}\right]\\ &=1+\frac16\left[\pi\csc\left(\frac\pi6\right)-6\right]\\ &=\frac\pi6\csc\left(\frac\pi6\right) \end{align}$$

$$\therefore~1-2\sum_{n=1}^\infty \frac{(-1)^n}{36n^2-1}=\frac\pi3$$

mrtaurho
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