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The series $\sum_{k=1}^{\infty }\frac{(-1)^{k+1}}{2k-1}=\frac{1}{1}-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\dots$ converges to $\frac{\pi}{4}$. Here, the sign alternates every term.

The series $\displaystyle\sum_{k=1}^{\infty }{(-1)^{\left(k^{2} + k + 2\right)/2} \over 2k-1}=\frac{1}{1}+\frac{1}{3}-\frac{1}{5}-\frac{1}{7}+\dots$ also converges. Here, the sign alternates every two terms.

What is the convergence value, explicitly, of the second series?

The first summation is noted above, because it might be a useful information to evaluate the second summation.

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    $\begingroup$ Wolfram says the result is $\frac{\sqrt{2}\pi}4$ $\endgroup$ – Peter Foreman Apr 17 at 10:28
  • $\begingroup$ @PeterForeman yes, but how to approach that value? $\endgroup$ – Hussain-Alqatari Apr 17 at 10:40
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Let $$\begin{align} S_1&=1-\frac15+\frac19-{1\over13}+\dots\\ S_2&=\frac13-\frac17+{1\over11}-{1\over15}+\dots \end{align}$$ so that the sum we seek is $S_1+S_2.$ T0 compute $S_1,$ consider $$f(x) = 1-{x^5\over5}+{x^9\over9}-{x^{13}\over13}+\dots$$ so that $$f'(x)=-x^4+x^8-x^{12}+\dots={-x^4\over1+x^4},\ |x|<1$$ and $$f(x)=\int_0^x{-t^4\over1+t^4}\mathrm{dt}+f(0),\ |x|<1$$ By Abel's limit theorem, $$S_1=\lim_{x\to1-}\int_0^x{-t^4\over1+t^4}\mathrm{dt}+f(0)=1-\int_0^1{t^4\over1+t^4}\mathrm{dt}$$ and we can do a similar calculation for $S_2$ to get $$S_2=\int_0^1{t^2\over1+t^4}\mathrm{dt}$$

The integrals are elementary, but tedious, and I leave them to you. (Frankly, I would do them by typing them into WolframAlpha. You can get the indefinite integrals, and check them by differentiation if you want to.) If you really want to do them by hand, I think it's easiest to consider only the definite integrals, and split the denominator into linear factors using complex numbers, but I don't know if you've learned about complex integrals yet.

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    $\begingroup$ How awesome is your answer?! Thank you very much. I got it. I will try to find the convergence value when the signs of the terms alternate every three terms. $\endgroup$ – Hussain-Alqatari Apr 17 at 12:56
  • $\begingroup$ Note that $S_1 = \int_0^1 \mathrm{dt} -\int_0^1{t^4\over1+t^4}\mathrm{dt} = \int_0^1{1\over1+t^4}\mathrm{dt}$, which might be marginally easier to integrate. $\endgroup$ – Michael Seifert Apr 17 at 14:11
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Hint:

$$\frac{1}{1}+\frac{1}{3}-\frac{1}{5}-\frac{1}{7}+\dots=\sum_{k=1}^{\infty }\frac{1}{8k-7}+\frac{1}{8k-5}-\frac{1}{8k-3}-\frac{1}{8k-1}$$

$$=\sum_{k=1}^{\infty }(\frac{1}{8k-7}-\frac{1}{8k-1})+(\frac{1}{8k-5}-\frac{1}{8k-3})$$

$$=\sum_{k=1}^{\infty }(1+\frac{1}{8k+1}-\frac{1}{8k-1})+(\frac{1}{3}+\frac{1}{8k+3}-\frac{1}{8k-3})$$

$$=\frac{4}{3}+\sum_{k=1}^{\infty }(\frac{-2}{64k^2-1}+\frac{-6}{64k^2-9})$$ $$=\frac{4}{3}-\frac{1}{32}\sum_{k=1}^{\infty }\frac{1}{k^2-\frac{1}{64}}-\frac{3}{32}\sum_{k=1}^{\infty }\frac{1}{k^2-\frac{9}{64}}$$

then use $$\frac{1-\pi x \cot(\pi x)}{2x^2}=\sum_{k=1}^{\infty }\frac{1}{k^2-x^2}$$

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$$ \begin{align} &\sum_{k=0}^\infty\left(\frac1{8k+1}+\frac1{8k+3}-\frac1{8k+5}-\frac1{8k+7}\right)\\ &=\sum_{k=0}^\infty\left(\frac1{8k+1}-\frac1{8k+7}\right)+\sum_{k=0}^\infty\left(\frac1{8k+3}-\frac1{8k+5}\right)\tag1\\ &=\sum_{k\in\mathbb{Z}}\frac1{8k+1}+\sum_{k\in\mathbb{Z}}\frac1{8k+3}\tag2\\ &=\frac18\sum_{k\in\mathbb{Z}}\frac1{k+\frac18}+\frac18\sum_{k\in\mathbb{Z}}\frac1{k+\frac38}\tag3\\ &=\frac\pi8\left[\cot\left(\frac\pi8\right)+\cot\left(\frac{3\pi}8\right)\right]\tag4\\[6pt] &=\frac{\pi\sqrt2}4\tag5 \end{align} $$ Explanation:
$(1)$: separate two absolutely convergent series
$(2)$: each series can be written as a sum over $\mathbb{Z}$
$(3)$: factor $\frac18$ out of each series
$(4)$: apply $(7)$ from this answer
$(5)$: evaluate; $\cot\left(\frac\pi8\right)=1+\sqrt2$ and $\cot\left(\frac{3\pi}8\right)=-1+\sqrt2$

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  • $\begingroup$ For sign change every $3$ terms, $\cot\left(\frac\pi{12}\right)=2+\sqrt3$, $\cot\left(\frac{3\pi}{12}\right)=1$, $\cot\left(\frac{5\pi}{12}\right)=2-\sqrt3$. $\endgroup$ – robjohn Apr 18 at 12:05
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[10px,#ffd]{{1 \over 1} + {1 \over 3} - {1 \over 5} - {1 \over 7} + {1 \over 9} + {1 \over 11} - \cdots} \equiv \sum_{n = 0}^{\infty}\pars{-1}^{n} \sum_{k = 2n + 1}^{2n + 2}{1 \over 2k - 1} \\[5mm] = & \sum_{n = 0}^{\infty}\pars{-1}^{n} \sum_{k = 0}^{1}{1 \over 2k + 4n + 1} = \sum_{k = 0}^{1}\sum_{n = 0}^{\infty} {\pars{-1}^{n} \over 4n + 2k + 1} \\[5mm] = &\ \sum_{k = 0}^{1}\sum_{n = 0}^{\infty}\pars{-1}^{n} \int_{0}^{1}t^{4n + 2k}\,\dd t = \sum_{k = 0}^{1}\int_{0}^{1}t^{2k} \sum_{n = 0}^{\infty}\pars{-t^{4}}^{n}\,\dd t \\[5mm] = & \sum_{k = 0}^{1}\int_{0}^{1}{t^{2k} \over 1 + t^{4}}\,\dd t = \sum_{k = 0}^{1}\int_{0}^{1}{t^{2k} - t^{2k + 4} \over 1 - t^{8}}\,\dd t \\[5mm] = &\ {1 \over 8}\sum_{k = 0}^{1}\int_{0}^{1} {t^{k/4 - 7/8} - t^{k/4 - 3/8} \over 1 - t}\,\dd t \\[5mm] = &\ {1 \over 8}\sum_{k = 0}^{1}\bracks{% \Psi\pars{{k \over 4} + {5 \over 8}} - \Psi\pars{{k \over 4} + {1 \over 8}}} \end{align}

where $\ds{\Psi}$ is the Digamma Function.

Then, \begin{align} &\bbox[10px,#ffd]{{1 \over 1} + {1 \over 3} - {1 \over 5} - {1 \over 7} + {1 \over 9} + {1 \over 11} - \cdots} \\[5mm] = &\ {\bracks{\Psi\pars{5/8} - \Psi\pars{1/8}} + \bracks{\Psi\pars{7/8} - \Psi\pars{3/8}} \over 8} \\[5mm] = &\ {\bracks{\Psi\pars{5/8} - \Psi\pars{3/8}} + \bracks{\Psi\pars{7/8} - \Psi\pars{1/8}} \over 8} \end{align}

With Euler Reflection Formula:

\begin{align} &\bbox[10px,#ffd]{{1 \over 1} + {1 \over 3} - {1 \over 5} - {1 \over 7} + {1 \over 9} + {1 \over 11} - \cdots} = {\pi\cot\pars{3\pi/8} + \pi\cot\pars{\pi/8} \over 8} \\[5mm] = &\ \pi\,{\tan\pars{\pi/8} + \cot\pars{\pi/8} \over 8} = {\pi \over 8\sin\pars{\pi/8}\cos\pars{\pi/8}} = {\pi \over 4\sin\pars{\pi/4}} \\[5mm] = &\ {\pi \over 4\pars{\root{2}/2}} = \bbx{{\root{2} \over 4}\,\pi} \approx 1.1107 \end{align}

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