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Find the value of the series: $$1-\frac{1}{2}-\frac{1}{4}-\frac{1}{6}-\frac{1}{8}+\frac{1}{3}-\frac{1}{10}-\frac{1}{12}-\frac{1}{14}-\frac{1}{16}+\frac{1}{5}\cdots$$

I know that the alternated harmonic series $$\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}$$ converges to $\ln2$, but here the order of the terms is different.

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  • $\begingroup$ The harmonic sum most certainly does not converge to $\ln 2$. $\endgroup$ – 5xum Jan 25 '16 at 11:00
  • $\begingroup$ With changing signs, I meant. (Leibniz) $\endgroup$ – Osh24 Jan 25 '16 at 11:02
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    $\begingroup$ Could you show your work on the problem? I think maybe it is being downvoted for this reason. $\endgroup$ – GPhys Jan 25 '16 at 11:04
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    $\begingroup$ Hint: Express the sum of the $5n$ first terms as a linear combination of the harmonic numbers $$H_k=\sum_{i=1}^k\frac1i,$$ for some well chosen values of $k$, then use the asymptotics $$H_k=\log k+\gamma+o(1)$$ to conclude. $\endgroup$ – Did Jan 25 '16 at 11:10
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    $\begingroup$ The solution is $0$. I recall reading about rearrangements of this type and the general solution is $$\ln (2)+\frac{1}{2}\ln \left(\frac{n}{m}\right)$$ Where, $n=1$ and $m=4$ for your rearrangement and $n$ and $m$ correspond to the length of the positive/negative term lengths. Unfortunately, I do not remember how it was proved. $\endgroup$ – GPhys Jan 25 '16 at 11:59
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This is a special case of a result by Fon Brown, L. O. Cannon, Joe Elich, and David G. Wright, On Rearrangements of the Alternating Harmonic Series, The College Mathematics Journal, Vol. 16, No. 2. (Mar., 1985), pp. 135-138.

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Let us elaborate on @Arthur idea by regrouping the terms five by five.

$$S=\sum_{k=0}^\infty \left(\frac{1}{2k+1}-\frac{1}{8k+2}-\frac{1}{8k+4}-\frac{1}{8k+6}-\frac{1}{8k+8}\right)$$

Rewriting $\frac{1}{2k+1}=\frac{4}{4(2k+1)}=\frac{4}{8k+4}$ simplifies the sum to four terms

$$S=\sum_{k=0}^\infty \left(-\frac{1}{8k+2}+\frac{3}{8k+4}-\frac{1}{8k+6}-\frac{1}{8k+8}\right)$$

that have all even denominator, so

$$S=-\frac{1}{2}\sum_{k=0}^\infty \left(\frac{1}{4k+1}-\frac{3}{4k+2}+\frac{1}{4k+3}+\frac{1}{4k+4}\right)$$

The relation

$$\int_0^1 x^ndx=\frac{1}{n+1} $$

changes the fractions into integrals $$S=-\frac{1}{2}\sum_{k=0}^\infty \int_0^1 \left(x^{4k}-3x^{4k+1}+x^{4k+2}+x^{4k+3}\right)dx$$

Now we may interchange the order of summation and integration and use the sum of the geometric progression

$$\sum_{k=0}^\infty cx^{ak+b} = \frac{cx^b}{1-x^a}$$

to obtain

$$\begin{align} -2S&=\int_0^1 \sum_{k=0}^\infty \left(x^{4k}-3x^{4k+1}+x^{4k+2}+x^{4k+3}\right)dx\\ &=\int_0^1 \frac {1-3x+x^2+x^3}{1-x^4} dx \\ &=\int_0^1 \frac {1-2x-x^2}{(1+x)(1+x^2)} dx \\ &=\int_0^1 \left( \frac{1}{1+x}-\frac{2x}{1+x^2}\right)dx\\ &=\left(\log(2)-\log(2)\right)=0\\ \end{align}$$

as in showing $ 1-\frac{3}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}-\frac{3}{6}+\frac{1}{7}+\frac{1}{8}+\cdots=0 $.

The procedure here is very similar to one of the four proofs Jack D'Aurizio gives to this question, which includes the one here.

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