Find the sum of $1-\frac17+\frac19-\frac1{15}+\frac1{17}-\frac1{23}+\frac1{25}-\dots$ 
Find the sum of $$1-\frac17+\frac19-\frac1{15}+\frac1{17}-\frac1{23}+\frac1{25}-\dots$$
a) $\dfrac{\pi}8(\sqrt2-1)$
b) $\dfrac{\pi}4(\sqrt2-1)$
c) $\dfrac{\pi}8(\sqrt2+1)$
d) $\dfrac{\pi}4(\sqrt2+1)$

I have tried a lot..  But i can't find any way to solve it,  plz help me.. Advance thanks to u
 A: Alternating series test gives the sum is convergent, so we can bracket off
\begin{align}
1-\sum_{k=1}^\infty\left(\frac1{8k-1}-\frac1{8k+1}\right)&=1-\sum_{k=1}^\infty\frac{2}{64k^2-1}\\
&=1+\frac{1}{32}\sum_{k=1}^\infty\frac{1}{(\frac18)^2-k^2}\\
&=\frac18\left(\frac1{1/8}+\frac{2}{8}\sum_{k=1}^\infty\frac{1}{(\frac18)^2-k^2}\right)\\
&=\frac18\pi\cot(\frac18\pi)\tag{$\dagger$}\label{eqn:cot}\\
&=\frac18\pi(1+\sqrt2)
\end{align}
where we used the Laurent series for cotangent in \eqref{eqn:cot}
$$
\pi\cot(\pi z)=\frac1z+2z\sum_{k=1}^\infty\frac{1}{z^2-n^2}.
$$
A: Here is an alternative solution in the case you do not know the cotangent series given in the amazing solution by user10354138.  Let $$f(z):=1-z^6+z^8-z^{14}+z^{16}-z^{22}+\ldots\text{ for }z\in\mathbb{C}\text{ such that }|z|<1\,.$$
Then, $(1-z^8)\,f(z)=1-z^6$, so
$$f(z)=\frac{1+z^2+z^4}{(1+z^2)(1+z^4)}=\frac{1}{2\,(1+z^2)}+\frac{1}{4\,\Biggl(\frac{1}{2}+\left(z+\frac{1}{\sqrt{2}}\right)^2\Biggr)}+\frac{1}{4\,\Biggl(\frac{1}{2}+\left(z-\frac{1}{\sqrt{2}}\right)^2\Biggr)}$$
for all $z\in\mathbb{C}$ with $|z|<1$.  That is, 
$$\int\,f(x)\,\text{d}x=\frac{1}{2}\,\arctan(x)+\frac{\arctan(\sqrt{2}\,x+1)+\arctan(\sqrt{2}\,x-1)}{2\sqrt{2}}+\text{constant}\,.$$
That is,
$$\int_0^1\,f(x)\,\text{d}x=\frac{1}{2}\,\left(\frac{\pi}{4}\right)+\frac{\left(\frac{3\pi}{8}-\frac{\pi}{4}\right)+\left(\frac{\pi}{8}+\frac{\pi}{4}\right)}{2\sqrt{2}}=\frac{\pi}{8}\,\big(1+\sqrt{2}\big)\,.$$
However, since the series representation of $f(z)$ compactly converges for $z\in\mathbb{C}$ such that $|z|<1$, we can then integrate $f(z)=1-z^6+z^8-z^{14}+\ldots$ term-by-term to get
$$\int_0^1\,f(x)\,\text{d}x=1-\frac{1}{7}+\frac{1}{8}-\frac{1}{15}+\frac{1}{17}-\frac{1}{23}+\ldots\,.$$
This shows that
$$1-\frac{1}{7}+\frac{1}{8}-\frac{1}{15}+\frac{1}{17}-\frac{1}{23}+\ldots=\frac{\pi}{8}\,\big(1+\sqrt{2}\big)\approx 0.948059449\,.$$
A: We are looking for
$$1+\sum_{k=1}^\infty \left(\frac1{8k+1}-\frac1{8k-1}\right)=1-\sum_{k=1}^\infty \frac{2}{64k^2-1}\approx 1-\frac1{32}\zeta(2)\approx \frac{\pi}{8}(\sqrt2+1)$$
A: Using Extended Harmonic Numbers
Using formula $(2)$ from this answer, we get
$$
H_{-1/8}=\left(1+\sqrt2\right)\frac\pi2-4\log(2)-\sqrt2\,\log\left(1+\sqrt2\right)
$$
and
$$
H_{-7/8}=-\left(1+\sqrt2\right)\frac\pi2-4\log(2)-\sqrt2\,\log\left(1+\sqrt2\right)
$$
Therefore,
$$
\begin{align}
\sum_{k=1}^\infty\left(\frac1{8k-7}-\frac1{8k-1}\right)
&=\frac18\sum_{k=1}^\infty\left(\frac1{k\vphantom{\frac18}}-\frac1{k-\frac18}\right)-\frac18\sum_{k=1}^\infty\left(\frac1{k\vphantom{\frac18}}-\frac1{k-\frac78}\right)\\
&=\frac18H_{-1/8}-\frac18H_{-7/8}\\[6pt]
&=\left(1+\sqrt2\right)\frac\pi8
\end{align}
$$

Using Conditionally Convergent Harmonic Series
Using formula $(7)$ from this answer, we get
$$
\begin{align}
\sum_{k=1}^\infty\left(\frac1{8k-7}-\frac1{8k-1}\right)
&=\frac18\sum_{k=1}^\infty\left(\frac1{k-\frac78}-\frac1{k-\frac18}\right)\\
&=\frac18\sum_{k=1}^\infty\left(\frac1{k-1+\frac18}+\frac1{-k+\frac18}\right)\\
&=\frac18\sum_{k\in\mathbb{Z}}\frac1{k+\frac18}\\[3pt]
&=\frac\pi8\cot\left(\frac\pi8\right)\\[6pt]
&=\left(1+\sqrt2\right)\frac\pi8
\end{align}
$$
A: The alternating sum is certainly less than $1$, which rules out ${\pi\over4}(\sqrt2+1)\gt{3\over4}(1.4+1)=1.8$. It's also greater than $1-{1\over7}={6\over7}$, which rules out ${\pi\over8}(\sqrt2-1)\lt{4\over8}(2-1)={1\over2}$ and ${\pi\over4}(\sqrt2-1)\lt{4\over4}(1.5-1)={1\over2}$. So if you assume that one of the options is correct, it can only be option c), ${\pi\over8}(\sqrt2+1)$.
