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Compute the series

$$\sum_{k=0}^{\infty}\left(\frac{1}{4k+1} - \frac{1}{4k+2}\right)$$

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@Gigili: I improved your suggested edit – user17762 May 24 '12 at 8:43
@ Marvis: the second answer is exactly the thing that proves the relation between Gregory's limit and harmonic numbers i supposed to be there. Great solution! – user 1618033 May 24 '12 at 11:03
Thanks. But be careful while doing such manipulations. As I have mentioned in the post, make sure to work with partial sums and in the end get it into a way so that each partial sum converges to a finite number. – user17762 May 24 '12 at 11:06
up vote 12 down vote accepted

One way is as follows. Note that \begin{align} \sum_{k=0}^{\infty} \left(\frac1{4k+1} - \frac1{4k+2} \right) & = \sum_{k=0}^{\infty }\int_{0}^{1} (t^{4k} - t^{4k+1})dt\\ & = \int_0^{1} \sum_{k=0}^{\infty} t^{4k} (1-t) dt\\ &= \int_0^1 (1-t) \dfrac{1}{1-t^4} dt\\ & = \int_0^1 \dfrac{dt}{(1+t^2)(1+t)} \end{align} Now use partial fractions to integrate. \begin{align} \int_0^1 \dfrac{dt}{(1+t^2)(1+t)} & = \frac12 \left(\int_0^1 \dfrac{dt}{1+t} + \int_0^1 \dfrac{1-t}{1+t^2} dt\right)\\ & = \dfrac12 \left. \left( \log(1+t) + \arctan(t) - \frac12 \log(1+t^2)\right) \right \vert_0^1\\ & = \frac12 \left(\log(2) + \frac{\pi}{4} - \frac12 \log(2) \right)\\ & = \frac12 \left( \frac{\log(2)}{2} + \frac{\pi}{4}\right)\\ & = \frac18 \left( \pi + \log(4) \right) \end{align}

Hence, we get that $$\sum_{k=0}^{\infty} \left(\frac1{4k+1} - \frac1{4k+2} \right) = \frac18 \left( \pi + \log(4) \right)$$


Here is another relatively round-about method which only uses the two identities v.i.z $$1 - \frac12 + \frac13 - \frac14 + \cdots = \log(2)$$ and the Madhava–Leibniz-Gregory series$$1 - \frac13 + \frac15 - \frac17 + \cdots = \dfrac{\pi}{4}$$ The idea is to rearrange the partial sums in a nice way. Warning Note that usually it is dangerous to rearrange series that are not absolutely convergent. Proper care must be taken to avoid errors like $\infty - \infty$. We need to carefully work with appropriate partial sums and in the end get it into a way so that each partial sum converges to a finite number.

If we let $$S_n = \sum_{k=0}^{n} \left(\frac1{4k+1} - \frac1{4k+2} \right),$$ then $$2S_n = \sum_{k=0}^{n} \left(\frac2{4k+1} - \frac2{4k+2} \right) = \sum_{k=0,2,4}^{2n} \left( \dfrac{2}{2k+1} - \dfrac{2}{2k+2}\right)\\ = \sum_{k=0,1}^{2n} \left( \dfrac{2}{2k+1} - \dfrac{2}{2k+2}\right) - \sum_{k=1,3}^{2n-1} \left( \dfrac{2}{2k+1} - \dfrac{2}{2k+2}\right)\\ = \underbrace{\sum_{k=0,1}^{2n} \left( \dfrac{1}{2k+1} - \dfrac{1}{2k+2}\right)}_{A_n} + \sum_{k=0,1}^{2n} \left( \dfrac{1}{2k+1} - \dfrac{1}{2k+2}\right) - \sum_{k=1,3}^{2n-1} \left( \dfrac{2}{2k+1} - \dfrac{2}{2k+2}\right)\\ = A_n + \sum_{k=0,2}^{2n} \left( \dfrac{1}{2k+1} - \dfrac{1}{2k+2}\right) + \sum_{k=1,3}^{2n-1} \left( \dfrac{1}{2k+1} - \dfrac{1}{2k+2}\right) - \sum_{k=1,3}^{2n-1} \left( \dfrac{2}{2k+1} - \dfrac{2}{2k+2}\right)\\ = A_n + \underbrace{\left(\sum_{k=0,2}^{2n} \dfrac{1}{2k+1} - \sum_{k=1,3}^{2n-1} \dfrac1{2k+1} \right)}_{B_n} - \sum_{k=0,2}^{2n} \dfrac{1}{2k+2} + \sum_{k=1,3}^{2n-1} \left( {\dfrac{1}{2k+1}} - \dfrac{1}{2k+2} - \dfrac1{2k+1} + \dfrac{2}{2k+2}\right)\\ = A_n + B_n - \underbrace{\left(\sum_{k=0,2}^{2n} \dfrac{1}{2k+2} - \sum_{k=1,3}^{2n-1} \dfrac1{2k+2} \right)}_{C_n}$$ Hence, we have that $$2S_n = A_n + B_n - C_n$$ where $$A_n = \sum_{k=0,1}^{2n} \left( \dfrac{1}{2k+1} - \dfrac{1}{2k+2}\right),$$ $$B_n = \left(\sum_{k=0,2}^{2n} \dfrac{1}{2k+1} - \sum_{k=1,3}^{2n-1} \dfrac1{2k+1} \right) = \sum_{k=0}^{n-1} \left( \frac1{4k+1} - \frac1{4k+3}\right) + \frac1{4n+1}$$ and $$C_n = \left(\sum_{k=0,2}^{2n} \dfrac{1}{2k+2} - \sum_{k=1,3}^{2n-1} \dfrac1{2k+2} \right) = \frac12 \sum_{k=0}^{n} \left(\frac1{2k+1} - \frac1{2k+2} \right) + \frac1{4n+4}$$ Now note that $$\lim_{n \rightarrow \infty} A_n = \log(2)$$ $$\lim_{n \rightarrow \infty} B_n = \dfrac{\pi}{4}$$ $$\lim_{n \rightarrow \infty} C_n = \dfrac{\log(2)}{2}$$ Since $\displaystyle \left \lvert \lim_{n \rightarrow \infty} A_n \right \rvert, \left \lvert \displaystyle \lim_{n \rightarrow \infty} B_n \right \rvert, \left \lvert \displaystyle \lim_{n \rightarrow \infty} C_n \right \rvert < \infty$, we get that $$\lim_{n \rightarrow \infty} 2S_n = \lim_{n \rightarrow \infty} A_n + \lim_{n \rightarrow \infty} B_n - \lim_{n \rightarrow \infty} C_n = \log(2) + \frac{\pi}{4} - \frac{\log(2)}{2}$$ Hence, we get that $$\sum_{k=0}^{\infty} \left(\frac1{4k+1} - \frac1{4k+2} \right) = \dfrac12 \left( \frac{\log(2)}{2} + \frac{\pi}{4} \right) = \dfrac18 \left(\pi + \log(4)\right)$$

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Thanks for your answer. – user 1618033 May 24 '12 at 8:36
"Madhava–Leibniz-Gregory" Yes! – Pedro Tamaroff May 25 '12 at 17:31

And more generally, if $a+b+c+d=0$, $$\sum_{k=0}^\infty \left(\dfrac{a}{4k+1} + \dfrac{b}{4k+2}+\dfrac{c}{4k+3}+\dfrac{d}{4k+4}\right) = \dfrac{a-c}{8} \pi + \dfrac{a+c-2d}{4} \ln(2) $$

EDIT: The generalization is that for positive integers $n$, if $\sum_{j=0}^{n-1} a_j = 0$ then $$ \sum_{k=1}^\infty \sum_{j=0}^{n-1} \dfrac{a_j}{nk-j} = -\frac{1}{n} \sum_\omega \sum_{j=0}^{n-1} a_j \omega^j \ln(1-\omega)$$ where the sum is over all $n$'th roots of unity except $1$.

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+1. Nice little formula to know! – user17762 May 25 '12 at 3:39
@Robert Israel: wow, i haven't seen such a formula before! Thanks. Your solution is very exciting. I'm struggling to figure out where did you take it from ... – user 1618033 May 25 '12 at 13:57
@Robert Israel: is there a more general form of that first formula in terms of using any positive integer number instead of "4"? Probably it may be easily proved with the integral way of Marvis. – user 1618033 May 25 '12 at 14:04

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