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I am trying to prove this. I used the telescoping method, but the problem is I need the first fraction to be $\frac{1}{4(n+1)}$ and I also tried to relate it to alternating Harmonic series which didn't work. Any hint would be greatly appreciated.$$\sum_{n=1}^{\infty}\left ( \frac{1}{4n+1}-\frac{1}{4n} \right )=\frac{1}{8}\left ( \pi-8+6\ln{2} \right )$$

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  • $\begingroup$ Wasn't this asked just a few hours ago, except with an image instead of markdown? $\endgroup$ Commented Jun 28, 2020 at 23:14
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    $\begingroup$ I love how many different approaches to this are represented in the answers. Plus they give hints at how the techniques have some underlying relationships to each other. $\endgroup$
    – aschepler
    Commented Jun 29, 2020 at 1:58

6 Answers 6

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This may be computed using

\begin{align}S&=\sum_{n=1}^\infty\left(\frac1{4n+1}-\frac1{4n}\right)\\&=\sum_{n=1}^\infty\int_0^1x^{4n}-x^{4n-1}~\mathrm dx\tag1\\&=\int_0^1\sum_{n=1}^\infty(x^{4n}-x^{4n-1})~\mathrm dx\tag2\\&=\int_0^1\frac{x^4-x^3}{1-x^4}~\mathrm dx\tag3\\&=\int_0^1\frac{-x^3(x-1)}{(x-1)(x+1)(x^2+1)}~\mathrm dx\tag4\\&=\frac12\int_0^1\frac1{x+1}+\frac{x+1}{x^2+1}-2~\mathrm dx\tag5\\&=\frac12\ln(x+1)+\frac14\ln(x^2+1)+\frac12\arctan(x)-x\bigg|_0^1\\&=\frac34\ln(2)+\frac\pi8-1\end{align}

where the steps are given by

$(1):~\dfrac1k=\int_0^1x^{k-1}~\mathrm dx$.

$(2):$ the tails of the sum converging to $0$ when integrated over.

$(3):$ the geometric series formula.

$(4):$ factoring the numerator and denominator.

$(5):$ partial fraction decomposition.

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If we call our sum $S$, then we see that

\begin{align*} S&=\sum_{n=1}^{\infty}\left(\frac{1}{4n+1}-\frac{1}{4n}\right)\\ &=-\frac{1}{4}\left(\sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{n+\frac{1}{4}}\right)-\gamma+\gamma\right) \end{align*}

where $\gamma$ is the Euler-Mascheroni constant. The reason that I have done this reformulation of the problem is that there is a well known function called the digamma function defined by

$$\psi(s)=\frac{\Gamma'(s)}{\Gamma(s)}$$

where $\Gamma(s)$ is the gamma function. It is well known that

$$\psi(s+1)=\sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{n+s}\right)-\gamma$$

and so we get now that

$$S=-\frac{1}{4}\left(\psi(5/4)+\gamma\right)$$

Gauss's formula relates values of the digamma function at rational numbers to a finite number of elementary functions, namely that

$$\psi\left(\frac{r}{m}\right)=-\gamma-\ln(2m)-\frac{\pi}{2}\cot\left(\frac{r\pi}{m}\right)+2\sum_{n=1}^{\lfloor (m-1)/2\rfloor}\cos\left(\frac{2\pi nr}{m}\right)\ln\left(\sin\left(\frac{\pi n}{m}\right)\right)$$

This formula is only valid for $r<m$, and so by using the reccurence relation

$$\psi(s+1)=\psi(s)+\frac{1}{s}$$

we get that

\begin{align*} \psi\left(\frac{5}{4}\right)&=\psi\left(\frac{1}{4}\right)+4\\ &=-\gamma-\ln\left(2\left(4\right)\right)-\frac{\pi}{2}\cot\left(\frac{\pi}{4}\right)+2\sum_{n=1}^{\lfloor(3)/2\rfloor}\cos\left(\frac{2\pi n}{4}\right)\ln\left(\sin\left(\frac{\pi n}{4}\right)\right)+4\\ &=-\gamma-\ln\left(8\right)-\frac{\pi}{2}\cot\left(\frac{\pi}{4}\right)+2\cos\left(\frac{\pi}{2}\right)\ln\left(\sin\left(\frac{\pi}{4}\right)\right)+4 \end{align*}

substituting for our basic trig values and using properties of natural logs this simplifies to

$$\psi\left(\frac{5}{4}\right)=-\gamma-3\ln\left(2\right)-\frac{\pi}{2}+4$$

Plugging this back into our main formula for $S$ in terms of the digamma function yields that

$$S=\frac{1}{8}\left(6\ln\left(2\right)+\pi-8\right)$$

which completes our calculation. In general, this method can be used to solve any sum in form

$$\sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{n+c}\right)$$

where $c$ is rational.

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Another approach is to use the Anti-difference concept.

Since the digamma function is defined as $$ \psi \left( z \right) = {d \over {dz}}\ln \Gamma \left( z \right) $$ its functional equation is $$ \Delta \psi \left( z \right) = \psi \left( {z + 1} \right) - \psi \left( z \right) = {d \over {dz}}\ln \left( {z\,\Gamma \left( z \right)} \right) - {d \over {dz}}\ln \left( {\Gamma \left( z \right)} \right) = {d \over {dz}}\ln z = {1 \over z} $$

It follows that $$ \sum\limits_{n = 1}^N {{1 \over {n + a}}} = \sum\limits_{n = 1}^N {\psi \left( {n + a + 1} \right) - \psi \left( {n + a} \right)} = \psi \left( {N + a + 1} \right) - \psi \left( {1 + a} \right) $$ and therefore $$ \sum\limits_{n = 1}^N {{1 \over {4n + 1}} - {1 \over {4n}}} = {1 \over 4}\sum\limits_{n = 1}^N {{1 \over {n + 1/4}} - {1 \over n}} = {1 \over 4}\left( {\psi \left( {N + 5/4} \right) - \psi \left( {5/4} \right) - \psi \left( {N + 1} \right) + \psi \left( 1 \right)} \right) $$

Since $\psi(z)$ is holomorphic for $0<\Re(z)$, then $$ \psi \left( {N + 5/4} \right) - \psi \left( {N + 1} \right) = \psi ^{\,\left( 1 \right)} \left( {N + 1} \right){1 \over 4} + {{\psi ^{\,\left( 2 \right)} \left( {N + 1} \right)} \over {2!}}\left( {{1 \over 4}} \right)^2 + \cdots $$ and since $$ \mathop {\lim }\limits_{N \to \infty } \psi ^{\,\left( k \right)} \left( {N + 1} \right) = 0\quad \left| {\;1 \le k} \right. $$ Therefore $$ \mathop {\lim }\limits_{N \to \infty } \sum\limits_{n = 1}^N {{1 \over {4n + 1}} - {1 \over {4n}}} = {1 \over 4}\left( {\psi \left( 1 \right) - \psi \left( {5/4} \right)} \right) = {3 \over 4}\ln 2 + {\pi \over 8} - 1 $$

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    $\begingroup$ Some of the terms used are divergent series (or "infinite antidifferences"), but it looks like this could be fixed up with some consideration for $\lim_{N \to \infty} \sum_{n=a}^{N+b}$ sums. $\endgroup$
    – aschepler
    Commented Jun 29, 2020 at 2:04
  • $\begingroup$ Most of the first half of the answer could (and probably should, if one wants to be rigorous and clear in notation) be ignored, and one could instead just use the fact that $\psi(a)-\psi(b)=\sum_{n\ge0}\left(\frac1{n+b}-\frac1{n+a}\right)$. $\endgroup$ Commented Jun 29, 2020 at 4:19
  • $\begingroup$ @aschepler I took your suggestion and put it in more rigorous terms $\endgroup$
    – G Cab
    Commented Jun 29, 2020 at 13:39
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This series can be evaluated with some generating function techniques. It will be related to the well-known generating function

$$ H(x) = \sum_{n=1}^\infty \frac{1}{n} x^n $$

which has closed form found by:

$$ H'(x) = \sum_{n=1}^\infty x^{n-1} = \sum_{n=0}^\infty x^n = \frac{1}{1-x} $$ $$ H(x) = \int\! \frac{dx}{1-x} = \ln(1-x) + C = \ln(1-x) $$

since $H(0)=0$ requires $C=0$. For this problem, we'll want $\ln$ to represent the principal branch of the complex function.

But the function $f$ involves only subsequences of $1/n$, not every term. A standard technique for dealing with subsequences in generating functions is a selector function for roots of unity:

$$ \sigma_k(x) = \frac{1}{k} \sum_{\ell=0}^{k-1} x^{\ell} $$

which has the property that for every $m \in \mathbb{Z}$,

$$ \sigma_k \! \left(e^{2 \pi m i/k}\right) = \begin{cases} 1 & \quad\mathrm{if}\ m \equiv 0 \pmod{k} \\ 0 & \quad\mathrm{if}\ m \not\equiv 0 \pmod{k} \end{cases} $$

To match the series terms, note that

$$ \forall m\in \mathbb{Z}: \sigma_4(i^{m-1}) - \sigma_4(i^m) = \begin{cases} 1 & \quad\mathrm{if}\ m \equiv 1 \pmod{4} \\ -1 & \quad\mathrm{if}\ m \equiv 0 \pmod{4} \\ 0 & \quad\mathrm{otherwise} \end{cases} $$

So define

$$ f(x) = \sum_{m=4}^\infty \frac{\sigma_4(x^{m-1}) - \sigma_4(x^m)}{m} $$

and the desired infinite sum will be $f(i)$.

$$ f(x) = -1 + \sum_{m=1}^\infty \frac{\sigma_4(x^{m-1}) - \sigma_4(x^m)}{m} $$

$$ f(x) = -1 + \sum_{m=1}^\infty \frac{1}{4m} (1 + x^{m-1} + x^{2m-2} + x^{3m-3} - 1 - x^m - x^{2m} - x^{3m}) $$

$$ f(x) = -1 + \frac{1}{4} \sum_{m=1}^\infty \left[ (x^{-1} - 1)\frac{x^m}{m} + (x^{-2} - 1)\frac{x^{2m}}{m} + (x^{-3} - 1)\frac{x^{3m}}{m} \right] $$

$$ f(x) = -1 + \frac{1}{4}(x^{-1} - 1)\ln(1-x) + \frac{1}{4}(x^{-2} - 1)\ln(1-x^2) + \frac{1}{4}(x^{-3} - 1)\ln(1-x^3) $$

Then the infinite sum is

$$ f(i) = -1 + \frac{-1-i}{4} \ln(1-i) - \frac{2}{4} \ln 2 + \frac{-1+i}{4}\ln(1+i) $$

Since $1-i = \sqrt{2}\,e^{-\pi i/4}$ and $1+i = \sqrt{2}\,e^{\pi i/4}$,

$$ f(i) = -1 + \frac{-1-i}{4}\left(\frac{1}{2} \ln 2 - \frac{\pi i}{4}\right) - 2 \ln 2 + \frac{-1+i}{4}\left(\frac{1}{2} \ln 2 + \frac{\pi i}{4}\right) $$

Multiplying out and collecting like terms cancels all the imaginary parts. (If they didn't cancel, we'd know something was wrong with the calculations, since the original sum is clearly real!) So finally, the sum is

$$ f(i) = -1 + \frac{\pi}{8} - \frac{3}{4} \ln 2 $$

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May be interesting to consider the most general case of $$S_p=\sum _{n=1}^{p } \left(\frac{1}{a n+b}-\frac{1}{c n+d}\right)$$ Using the digamma function, $$S_p=\frac{c \psi \left(\frac{b}{a}+p+1\right)-a \psi \left(\frac{d}{c}+p+1\right)-c \psi \left(\frac{b}{a}+1\right)+a \psi\left(\frac{d}{c}+1\right)}{a c}$$ Expanded as series for large values of $p$ $$S_p=\frac{(c-a) \log (p)-c \psi \left(\frac{a+b}{a}\right)+a \psi \left(\frac{c+d}{c}\right)}{a c}+O\left(\frac{1}{p}\right)$$ which can converge only if $c=a$. In such a case $$S_p=\frac{\psi \left(\frac{a+d}{a}\right)-\psi \left(\frac{a+b}{a}\right)}{a}+O\left(\frac{1}{p}\right)$$ For the particular case where $d=0$, $\psi(1)=-\gamma$ and $$S_p=-\frac 1a\left(\psi \left(1+\frac{b}{a}\right)+\gamma \right)+O\left(\frac{1}{p}\right)$$ and if $a=k b$ the term in parentheses has simple expressions up to ... $k=4$.

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We don't need anything fancy. Here $N=4$.

If $a_n$ is $N$-periodic and $\sum_{n=1}^N a_n = 0$ then $$\sum_{n=1}^\infty \frac{a_n}{n} = \sum_{n=1}^\infty \frac{\frac1N \sum_{k=1}^{N-1} A_k e^{2i\pi nk/N}}{n}=-\frac1N \sum_{k=1}^{N-1} A_k \log(1-e^{2i\pi k/N})$$ where $A_k=\sum_{n=1}^N a_n e^{-2i\pi nk/N}$

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