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I would like to sum the following $$\sum_{n=1}^{\infty}\frac{1}{n^b(n+m)^b}$$ for values of $b$ for which the sum exists. The problem is that truncating the sum does not work that good, and needs in general large number of terms specially when $b$ is close to non-summability areas. I then tried to rewrite the sum into some function for which we might have convenient and precise ways to calculate them. Unfortunately have not been successful. I appreciate any ideas. Many thanks!

Math-fun
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    If $b$ is an integer, you may compute such series through partial fraction decomposition and the functions $\psi,\psi',\psi'',\ldots$ where $\psi(x)=\frac{d}{dx}\log\Gamma(x)$. – Jack D'Aurizio Dec 15 '16 at 14:48
  • That's the problem. $b $ is not an integer. – Math-fun Dec 15 '16 at 14:50
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    The apply the inverse Laplace transform to convert your series into an integral. – Jack D'Aurizio Dec 15 '16 at 14:50
  • That sounds interesting I will try. Thank you :-) – Math-fun Dec 15 '16 at 14:51
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    In the critical case $b\approx\frac{1}{2}$, you will see that the behaviour of your series depends on the behaviour of a Bessel $I$ function. – Jack D'Aurizio Dec 15 '16 at 14:52
  • The thing is observing the link to bessel functions I thought there might be some functions still related that I don't know. But I understand the direction of your suggestion. – Math-fun Dec 15 '16 at 14:54
  • I took the liberty of putting the formula in the title, which (I think) makes the question a little easier to browse. You can revert it if it is not desired. But you might want to proofread the rest of the question as well to make it easier to read. – David K Dec 15 '16 at 15:06
  • @DavidK this is great help. Thank you very much. I made some edits, sorry for the bad text. – Math-fun Dec 15 '16 at 15:13
  • @JackD'Aurizio Jack, first Happy Holidays! As far as the first comment, suppose $b=k\in \mathbb{N}$. I don't see a direct connection between the derivatives of the digamma function, $$\psi^{(k)}(m)=(-1)^{k+1},k!,\sum_{n=1}^\infty\frac{1}{(n+m)^{k+1}} \tag 1$$and the function of interest herein, $$f_k(m)=\frac{1}{m^k}\sum_{n=1}^\infty \left(\frac{1}{n}-\frac{1}{n+m}\right)^k \tag 2$$There seems to be a lot of work if we can make use of $(1)$ to express $(2)$. If $b$ in not an integer, I don't see how application of the inverse Laplace transform converts the sum to an integral. – Mark Viola Dec 15 '16 at 19:06
  • @Dr.MV: if you write $\frac{1}{n(n+m)}$ as $\frac{1}{m}\left(\frac{1}{n}-\frac{1}{n+m}\right)$, then consider the $b$-th power, you easily get terms manageable through partial fraction decomposition / the inverse Laplace transform / derivatives of the digamma function. Happy holidays to you too! – Jack D'Aurizio Dec 15 '16 at 19:39
  • @JackD'Aurizio Jack, as embedded in my comment, the $k$'th power, $\left(\frac1n - \frac{1}{n+m}\right)^k$ involve terms $\frac{1}{n^\ell}\frac{1}{(n+m)^{k-\ell}}$ for $\ell = 0 , \dots, k$. Then, how does one proceed manageably? – Mark Viola Dec 15 '16 at 20:13
  • @Dr.MV: to state it clearly, $$\frac{1}{n^b(n+m)^b}=\frac{\sqrt{\pi}}{\Gamma(b)}\int_{0}^{+\infty}e^{-ns}e^{-ms/2}(m/s)^{1/2-b} I_{b-1/2}(ms/2),ds. $$ – Jack D'Aurizio Dec 15 '16 at 20:39
  • @JackD'Aurizio Jack, thanks for this. Did you derive this expression or is this a "look up" in a treatise such as Gradshteyn? It's not obvious to me how one starts with the LHS and arrives at the RHS integral representation. Finally, given this relationship, the rest is easy. In summing the LHS for $b>1/2$, we can justify easily interchanging the integral and the series (e.g., exploit uniform convergence for $b\ge b^{*} >1/2$) and sum the geometric series. – Mark Viola Dec 15 '16 at 21:45
  • @Dr.MV: kind of both. I started computing the inverse Laplace transform by hand, got it wrong a couple of times, convinced myself it has to involve the Bessel I function, invoked greater gods for a confirmation :D – Jack D'Aurizio Dec 15 '16 at 21:58
  • @JackD'Aurizio What are you viewing as the transform variable for $\frac{1}{n^b,(n+m)^b}$? – Mark Viola Dec 15 '16 at 22:00
  • @Dr.MV: $n$, in order to convert the series into $\int_{0}^{+\infty}\sum_{n\geq 1}e^{-ns}\omega(s),ds.$ – Jack D'Aurizio Dec 15 '16 at 22:04
  • So, you attempted to evaluate $$\frac{1}{2\pi i} \int_{\sigma - i\infty }^{\sigma + i\infty} \left(\frac{1}{t^b(t+m)^b}\right),e^{st},dt?$$ – Mark Viola Dec 15 '16 at 23:06
  • If you use a Mellin transform with respect to $m$ you can get the original sum $S(b,m)$ $$ S(b,m)=\frac{1}{2\pi i} \int_{\sigma-i\infty}^{\sigma + i\infty} \frac{\Gamma(b-s)\Gamma(s)\zeta(2b-s)}{\Gamma(b)m^s} ; ds $$ – Benedict W. J. Irwin Jun 12 '17 at 16:10
  • For integer values of $b$, a closed form is given here: https://math.stackexchange.com/q/5010496 – Gary Feb 04 '25 at 08:42

1 Answers1

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We know that, if $b>\frac 12$ and $m>0$, $$\int\frac{dn}{n^b(n+m)^b}=-\frac{n^{1-b} (m+n)^{1-b}}{(b-1) m}\, \, _2F_1\left(1,2-2 b;2-b;-\frac{n}{m}\right)$$ $$\int_{1}^{\infty}\frac{dn}{n^b(n+m)^b}=\frac 1{2b-1}\,\, _2F_1(b,2 b-1;2 b;-m)$$

Using the simplest form of Euler-MacLaurin summation $$\sum_{n=1}^{\infty}\frac{1}{n^b(n+m)^b}=\frac{(m+1)^{1-b}}{(b-1) m} \, \,_2F_1\left(1,2-2 b;2-b;-\frac{1}{m}\right)+$$ $$m^{1-2 b}\,\,\frac{ \Gamma (1-b)\,\, \Gamma (2 b-1)}{\Gamma (b)}- \frac{(m+1)^{-b-7}}{1209600}\sum_{p=0}^7 A_p(m)\,b^p$$ where the coefficients are

$$\left( \begin{array}{cc} p & A_p(m) \\ 0 & -604800 (m+1)^7 \\ 1 & -240 (m+2) \left(407 m^6+2455 m^5+6157 m^4+8217 m^3+6155 m^2+2457 m+407\right) \\ 2 & 4 (m+2) \left(1201 m^6+7175 m^5+19077 m^4+28526 m^3+24708 m^2+11246 m+2402\right) \\ 3 & 2 (m+2) \left(952 m^6+7139 m^5+22285 m^4+36460 m^3+32710 m^2+15344 m+3808\right) \\ 4 & 5 (m+2) \left(67 m^6+410 m^5+1074 m^4+1544 m^3+1392 m^2+968 m+536\right) \\ 5 & 5 (m+2)^3 \left(27 m^4+106 m^3+190 m^2+184 m+108\right) \\ 6 & 21 (m+2)^5 \left(m^2+2 m+2\right) \\ 7 & (m+2)^7 \\ \end{array} \right)$$

which provide decent approximations.

Claude Leibovici
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