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How to show that the following infinite series $$ 1-\frac{3}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}-\frac{3}{6}+\frac{1}{7}+\frac{1}{8}+\cdots=0? $$ The above series is of the form $\sum_{n \ge 1} \frac{f(n)}{n}$, where $f$ is a periodic arithmetical function of period $4$, with the values $f(1)=f(3)=f(4)=1$ and $f(2)=-3$. Since $\sum_{1 \le i \le 4} f(i)=0$, it is assured that this series is convergent.

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I'm a little wary of doing this, since it's not absolutely convergent, but can you rewrite each period as

$$ \left(\frac{1}{4k-3}-\frac{1}{4k-2}+\frac{1}{4k-1}-\frac{1}{4k}\right) - \left(\frac{2}{4k-2}-\frac{2}{4k}\right) $$

which can in turn be rewritten as

$$ \left(\frac{1}{4k-3}-\frac{1}{4k-2}+\frac{1}{4k-1}-\frac{1}{4k}\right) - \left(\frac{1}{2k-1}-\frac{1}{2k}\right) $$

and then you have two parallel alternating harmonic series, both of which sum to $\ln 2$. Subtract one from the other and you get $0$.

But, as I say, I'm not sure that's kosher.

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  • $\begingroup$ The manipulation should also be valid on the $4k$th partial sum, and get $S_{4k} - S_{2k}$ where $S_n$ is the $n$th partial sum of the $\ln 2$ series. $\endgroup$ – Daniel Schepler May 10 '17 at 22:30
  • $\begingroup$ It is a valid technique. For $n=4m+r,$ with $r\in \{0,1,2,3\}$ you are taking $\sum_{i=0}^nA_i=(\sum_{j=0}^m \sum_{k=0}^3 A_{4j+k})+ (\sum_{l=1}^rA_{4m+l}).$.... The expression $\sum_{l=1}^r A_{4m+l}$ has $0,1,2$,or $3$ terms in it , and it $\to 0$ as $n\to \infty.$ The rest is equal to $(B_m+\ln 2)-(C_m+\ln 2)$ where $B_m$ and $C_m$ go to $0$ as $n\to \infty.$ $\endgroup$ – DanielWainfleet May 12 '17 at 4:52
  • $\begingroup$ @DanielWainfleet: Thanks, yes, I had to sit down and think about it after I wrote it. $\endgroup$ – Brian Tung May 12 '17 at 17:13
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Your series is

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

and this zero relation was the motivation for OEIS sequence http://oeis.org/A176563

It can also be constructed from $2\log(2)-\log(4)$ as follows

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

For the formula for $\log(4)$, see Do these series converge to logarithms?

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sum of all odd terms: $X = (1 + 1/3 + 1/5 + ... )$

sum of all terms (4n+2): $Y = -3/2(1 + 1/3 + 1/5 + ...) = -3X/2$

Sum of remaining terms = $Z = 1/4 + 1/8 + 1/12 + ... = 1/4(1 + 1/2 + 1/3 + 1/4 + 1/5 + ...)$

splitting into odd and even terms: $Z = 1/4(1+ 1/3 + 1/5 +...) + 1/8(1 + 1/2 + 1/3 + ...)$

So $Z = X/4 + Z/2 \implies Z = X/2$

So $X + Y + Z = X - 3X/2 + X/2 = 0$

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