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Recently, there was some discussion $[1]$, $[2]$ on certain infinite sums, like these

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

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

$$h_0=\sum_{k=1}^\infty \frac{\sin(k)}{k}$$

$$h_1=\sum_{k=1}^\infty \frac{\sin(k^2)}{k}$$

$$h_2=\sum_{k=1}^\infty \frac{\sin(k^2)}{k^2}$$

and I wondered if the elegant summation formula of Abel and Plana could be applied to these sums. In summary, AP works fine for $f$, $g$, and $h_0$ but does not for $h_1$ and $h_2$.

This led me to the question in the heading.

But let us start from the beginning.

Quoting $[3]$: "In mathematics, the Abel-Plana formula (AP) is a summation formula discovered independently by Niels Henrik Abel (1823) and Giovanni Antonio Amedeo Plana (1820). It states that

$$\sum _{n=0}^{\infty } f(n)=\frac{1}{2}f(0)+\int_0^{\infty } f(x) \, dx+i \int_0^{\infty } \frac{f(i y)-f(-i y)}{e^{2 \pi y}-1} \, dy\tag{1}$$

It holds for functions f that are holomorphic in the region Re(z) >= 0, and satisfy a suitable growth condition in this region; for example it is enough to assume that |f| is bounded by $C/|z|^{1+\epsilon}$ in this region for some constants C, $\epsilon > 0$, though the formula also holds under much weaker bounds."

Ok, the growth condition for applicability of AP,

$$|f(z)| < C / |z|^{1+\epsilon}\text{, with } C>0, \epsilon >0\tag{2}$$

is fulfilled for $f$ and $g$ above, and is violated by the $h$-sums because $\sin(z)$ is not limited in the complex plane.

I have obtained the following results: AP works for $f$, $g$, and $h_0$ and doesn't for the remaining $h_{1,2}$. The surprising case is $h_0$ where the growth condition $(2)$ not met but AP still gives the correct result.

Examples

Example $g$. -> sucess

The sum has a closed form

$$g = \frac{1}{2} (\pi \coth (\pi )-1)\simeq 1.0766740474685811741$$

Now we to AP: Here $f(n)=\frac{1}{1+(n+1)^2}$ to let the index start at $n=0$. $f(z)$ is meromorhpic (it has two simple poles at $-1+\pm i$) and it goes to zero like $1/|z|^2 for $|z|\to\infty$ so that condition $(2)$ is fulfilled.

We write the r.h.s. of $(1)$ as a sum of three parts, a, b, and c, and calculate

$$a = \frac{1}{2}f(0)= \frac{1}{2}= 0.5$$ $$b = \int_0^{\infty } f(x) \, dx=\frac{\pi}{4}\simeq 0.785398$$

the integrand of the last integral is

$$c_i = i \frac{f(i y)-f(-i y)}{e^{2 \pi y}-1} = \frac{i (f(i y)-f(-i y))}{e^{2 \pi y}-1}= \frac{4 y}{\left(e^{2 \pi y}-1\right) \left(y^4+4\right)}$$

hence the integral is

$$c = \int_0^\infty \frac{4 y}{\left(e^{2 \pi y}-1\right) \left(y^4+4\right)}$$

The integral is convergent and the numerical value is

$$c_n = 0.041275695056277776209$$

Hence $a+b+c$ is equivalent to the closed expression for $g$. Success!

Example $h_0$ -> success

$$f(n) = \frac{\sin(n+1)}{n+1}$$

The sum has the closed form

$$\frac{1}{2} (\pi -1)\simeq 1.0707963267948966192$$

Condition $(2)$ is not fulfilled because for complex $z$ we have $f(z) \sim \sinh(|z|)$. Let us nevertheless apply AP.

$$a = \frac{1}{2}f(0)=\frac{\sin (1)}{2}\simeq 0.420735$$ $$b = \int_0^{\infty } f(x) \, dx=\frac{1}{2} (\pi -2 \text{Si}(1))\simeq 0.624713$$ $$c_i = i \frac{f(i y)-f(-i y)}{e^{2 \pi y}-1} = \frac{2 (y \sin (1) \cosh (y)-\cos (1) \sinh (y))}{\left(e^{2 \pi y}-1\right) \left(y^2+1\right)}$$ $$c = \int_0^\infty \frac{2 (y \sin (1) \cosh (y)-\cos (1) \sinh (y))}{\left(e^{2 \pi y}-1\right) \left(y^2+1\right)}$$

The integral is convergent and the numerical value is

$$c_n = 0.0253475$$

Hence we have $a+b+c \simeq 1.070796$ in good agreement with $h_0$. So in spite of violating the growing condition AP gives the correct result.

Example $h_1$ -> failure

$$f(n) = \frac{\sin((n+1)^2)}{n+1}$$

Numerically we have

$$h_1 \simeq 0.167924$$

Condition $(2)$ is not fulfilled because for complex $z$ we have $f(z) \sim \sinh(|z|^2)$. Let us nevertheless formally apply AP.

$$a = \frac{1}{2}f(0)=\frac{\sin (1)}{2}\simeq 0.420735$$ $$b = \int_0^{\infty } f(x) \, dx=\frac{1}{4} (\pi -2 \text{Si}(1))\simeq 0.312357$$ $$c_i = i \frac{f(i y)-f(-i y)}{e^{2 \pi y}-1} = \frac{\left(-(y+i) \sin \left((y-i)^2\right)-(y-i) \sin \left((y+i)^2\right)\right) (\coth (\pi y)-1)}{2 \left(y^2+1\right)}$$ $$c = \int_0^\infty c_i \,dy\simeq -0.0304174$$

and we have

$a+b+c \simeq 0.758439693815265076542$ whic siginifantly deviates from the value of the sum $h_1$.

References

$[1]$ Find Harmonic Numbers for Imaginary and Complex Values
$[2]$ Convergence of $\sum_{k=1}^\infty \frac{\sin(k(k-1))}{k}$
$[3]$ https://en.wikipedia.org/wiki/Abel%E2%80%93Plana_formula

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  • $\begingroup$ $g$ has poles at $\pm i$. $\endgroup$ – Daniel Fischer Dec 11 '19 at 17:00
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The proof on wikipedia is clear on the condition you need :

  • $\int_C \frac{f(z)}{e^{2i\pi z}-1}dz$ converges with $C$ enclosing $[a,\infty)$

  • $f$ is analytic there and it decays fast enough so that the residue theorem applies

  • split $C$ in two parts, replace $\frac1{e^{2i\pi z}-1}$ by $1-\frac1{e^{-2i\pi z}-1}$ in one of them and shift the contours (using that $g_1=\frac{f}{e^{2i\pi z}-1},g_2=\frac{f}{e^{-2i\pi z}-1}$ are analytic away from the real axis, or meromorphic with finitely many poles -in that case add the relevant residues- and they decay fast enough)

obtaining $$\sum Res(\frac{f(z)}{e^{2i\pi z}-1})=\int_{a-i\infty}^a\frac{f(z)}{e^{2i\pi z}-1}dz-\int_a^{a+i\infty}\frac{f(z)}{e^{-2i\pi z}-1}dz+\int_a^\infty f(z)dz\\+2i\pi\sum Res(g_1)-2i\pi\sum Res(g_2)$$

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  • $\begingroup$ @ reuns Thank you. Ok, I've seen that. But what happens specifically with $g$? $\endgroup$ – Dr. Wolfgang Hintze Dec 11 '19 at 18:41
  • $\begingroup$ With $1/(z^2+1)$ ? It happens what I wrote, there are two more residues at $\pm i$. If you don't understand the residue theorem and the Cauchy integral theorem, ie. the proof on wikipedia, then forget about Abel-Plana; $\endgroup$ – reuns Dec 11 '19 at 18:43
  • $\begingroup$ @ reuns Thank you for your clear language. Rest assured I know contour integration fairly well. But, unfortunately this knowledge does not prevent calculation errors. And that's what applies here. I have corrected my text and extended it with expamples. This leaves the suprising case $h_0$ where $f$ increases strongly in the complex and still AP gives the correct result. This surprise unfortunately does not repeat for $h_1$. $\endgroup$ – Dr. Wolfgang Hintze Dec 11 '19 at 20:06
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    $\begingroup$ That it works for $\sin z/ z$ and that it fails for $\sin(z^2)/z$ is not surprising, $\frac{\sin(ix)/x}{e^{2\pi |x|}-1}$ is fast decaying whereas $\frac{\sin(ix^2)/x}{e^{2\pi|x|}-1}$ is not. Follow the proof and see what is needed in each step. $\endgroup$ – reuns Dec 11 '19 at 20:11
  • $\begingroup$ @ reuns Thanks for the valuable hints with which I can answer the question myself. But I think it is a legitimate question to ask for the conditions of applicability of a certain formula, and that it is not an answer to say, as you did, "If you don't understand the residue theorem and the Cauchy integral theorem, ie. the proof on wikipedia, then forget about Abel-Plana". $\endgroup$ – Dr. Wolfgang Hintze Dec 12 '19 at 2:47

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