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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

Community
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user532616
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5 Answers5

5

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} $$ \pi\cot(\pi z)=\frac1z+2z\sum_{k=1}^\infty\frac{1}{z^2-n^2}. $$

user10354138
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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 $$1-\frac{1}{7}+\frac{1}{8}-\frac{1}{15}+\frac{1}{17}-\frac{1}{23}+\ldots=\frac{\pi}{8}\,\big(1+\sqrt{2}\big)\approx 0.948059449\,.$$

Batominovski
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We are looking for

$$1+\sum_{k=1}^\infty \left(\frac1{8k+1}-\frac1{8k-1}\right)=1-\sum_{k=1}^\infty \frac{2}{64k^2-1}\approx 1-\frac1{32}\zeta(2)\approx \frac{\pi}{8}(\sqrt2+1)$$

user
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  • We need to sum...I add one more step – user Nov 13 '18 at 07:13
  • Sir i have also reached there, but after that i don't know how to go ahead, plz help me by solving this, i shall be very thankful to u – user532616 Nov 13 '18 at 07:14
  • What about $$\sum_{k=1}^\infty \frac{2}{64k^2-1}\approx (1/32) \sum_{k=1}^\infty\frac1{k^2}$$ – user Nov 13 '18 at 07:19
  • Since you are given a multiple choice I think that an estimation can suffice to select the correct option (which should be the “a”). – user Nov 13 '18 at 07:34
  • After the editing for the multiple choices the correct aswer should be “d”. – user Nov 13 '18 at 08:49
  • I used the integral proposed by @Lord Shark The Unknown in Mathematica which gave the result $$\frac{\pi (1 +\sqrt 2)}{8}$$ (i.e choice "c") – Vectorizer Nov 18 '18 at 19:27
  • @Vectorizer Yes of course sorry it was a typo! Also my estimation agree with that! – user Nov 18 '18 at 19:30
  • $\zeta(2)=\frac{\pi^2}6$. What happens to the second factor of $\pi$? – robjohn Nov 22 '18 at 19:43
  • @robjohn my (brutal) evaluation, aimed to select the correct option, is solely numerical, that is $\approx 0.948$. – user Nov 22 '18 at 19:49
  • When I first read the question on my phone, I missed the options. Approximations work if you assume one of the options is correct. – robjohn Nov 22 '18 at 22:35
  • @robjohn Yes indeed my idea was that the multiple choice was probably aimed to a rough fast estimation instead of a rigorous proof for the exact value. Moreover I wasn't able to find that :) – user Nov 22 '18 at 22:43
1

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 $$ \begin{align} \sum_{k=1}^\infty\left(\frac1{8k-7}-\frac1{8k-1}\right) &=\frac18\sum_{k=1}^\infty\left(\frac1{k-\frac78}-\frac1{k-\frac18}\right)\\ &=\frac18\sum_{k=1}^\infty\left(\frac1{k-1+\frac18}+\frac1{-k+\frac18}\right)\\ &=\frac18\sum_{k\in\mathbb{Z}}\frac1{k+\frac18}\\[3pt] &=\frac\pi8\cot\left(\frac\pi8\right)\\[6pt] &=\left(1+\sqrt2\right)\frac\pi8 \end{align} $$

robjohn
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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), ${\pi\over8}(\sqrt2+1)$.

Barry Cipra
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