Consider the infinite Sum

$S=\sum\limits_{n\ \text{odd}}^{\infty}\left(\dfrac{1}{nt}\right)^2\left[1-i\left(nt\right)^2\right]^{-1}$

Is there a way to approximate this sum as a contour integral? In my physical problem it is valid to make the approximation $t\to 0$ in a certain regime (or more precisely, the approximation that $t$ is very small). So I tried the following:

Let us set aside the fact (for now) that this is a sum over odd terms only. So we rewrite the sum as


$\ \ =\dfrac{1}{t}\lim\limits_{t\to 0}\sum\limits_{n=1}^{\infty}t\cdot\left(\dfrac{1}{nt}\right)^2\left[1-i\left(nt\right)^2\right]^{-1}$

$\ \ =\dfrac{1}{t}\int\limits_0^\infty\dfrac{1}{x^2}\cdot\dfrac{1}{1-ix^2}\ dx$

(i.e.) setting $t\to dx$ and $nt\to x$ as this resembles a Riemann sum. But this doesn't work because the integral does not converge. Is there a nice way to approximate this infinite sum? Ideally, it'll be great if there is a way to do this without assuming $t$ is small.

EDIT: Corrected the problem statement based on Blumenthal's comment.

  • $\begingroup$ There's something funny with the units, so to speak. Even before dealing with the problem of a sum over odd $n$, what you'd have in the limit $t \rightarrow 0$ is that $t \rightarrow dx$ and $n \cdot t \rightarrow x$, not $n/t \rightarrow x$. Do you understand why? $\endgroup$ Commented Apr 29, 2013 at 21:59
  • $\begingroup$ I have non-dimensionalized all the variables in the problem. So $t$ here is dimensionless. It is not time. Sorry if this was confusing! I think the sum as stated in the first equation looks fine? $\endgroup$ Commented Apr 29, 2013 at 22:25
  • $\begingroup$ I'm speaking loosely when I say 'units'. My point was that if $t$ is supposed to become infinitesimal, then you should expect $n \cdot t \rightarrow x$, not the other way. The problem here is that to even begin a Riemann Sum argument for a summand over a set (here, it's $\mathbb{N}$), for instance, you need that $\{t n \mid n \in \mathbb{N} \}$ serves as a 'fine enough' mesh for the positive reals in the limit at $t \rightarrow 0$. The set $\{n / t \mid n \in \mathbb{N}\}$ flies off to infinity as $t \rightarrow 0$, and will be useless as a mesh. $\endgroup$ Commented Apr 29, 2013 at 22:35
  • $\begingroup$ Are you summing over positive $n$ only? $\endgroup$ Commented Apr 29, 2013 at 22:37
  • $\begingroup$ @ABlumenthal Yes, you are correct! I see your point --- I have corrected the problem statement now. Indeed,looking at the physics of my problem, the actual useful limit is $t\to 0$ as defined in the corrected statement. $\endgroup$ Commented Apr 29, 2013 at 23:16

3 Answers 3


(This is an answer to a previous version of the question.)

You can express your sum as a power series in $t$. First rewrite your sum as

$$ S = i \sum_{n = 0}^{\infty} \frac{\left(\frac{t}{2n+1}\right)^4}{1+i\left(\frac{t}{2n+1}\right)^2}. $$

For $|t| < 1$ and $n\geq 0$ we have

$$ \frac{1}{1+i\left(\frac{t}{2n+1}\right)^2} = \sum_{m=0}^{\infty} (-i)^m \left(\frac{t}{2n+1}\right)^{2m}, $$

so that, by changing the order of summation,

$$ \begin{align} S &= i \sum_{n = 0}^{\infty} \left(\frac{t}{2n+1}\right)^4 \sum_{m=0}^{\infty} (-i)^m \left(\frac{t}{2n+1}\right)^{2m+4} \\ &= i \sum_{m=0}^{\infty} (-i)^m t^{2m+4} \sum_{n=0}^{\infty} \frac{1}{(2n+1)^{2m+4}} \\ &= i \sum_{m=0}^{\infty} (-i)^m (1-4^{-m-2}) \zeta(2m+4) t^{2m+4}, \end{align} $$

where we have used the identity

$$ \sum_{n=0}^{\infty} \frac{1}{(2n+1)^\alpha} = (1-2^{-\alpha})\zeta(\alpha). $$

It follows, for example, that

$$ S = i \frac{\pi^4}{96} t^4 + \frac{\pi^6}{960} t^6 + O(t^8) $$

as $t \to 0$.

  • $\begingroup$ Thanks, seems like this might work for me. I just corrected the problem statement based on what @ABlumenthal pointed out. I guess, your solution still works be replacing $t$ with $1/t$ $\endgroup$ Commented Apr 29, 2013 at 23:25
  • $\begingroup$ It'll be a great help if you can modify your solution based on the corrected problem statement. $\endgroup$ Commented Apr 29, 2013 at 23:36
  • $\begingroup$ @G.H.Hardly, I don't see an easy way to adapt my method to the current problem statement. It might be better to un-accept my answer in the hopes that someone else can answer your revised question. $\endgroup$ Commented Apr 30, 2013 at 2:44
  • $\begingroup$ @AntonioVargas Your method might still work, but the complication is that your series expansion for the geometric sum is no longer valid for huge $n$. I'm guessing the procedure here would be to fix $t$ small and find some estimate for the tail of the sum (here 'tail' means $n > 1/t$). $\endgroup$ Commented Apr 30, 2013 at 16:54
  • $\begingroup$ @AntonioVargas On second thought, the tail $\{n > 1/t\}$ will resemble some kind of Riemann sum of an integrable function (since the bounds of integration are now $1$ to $\infty$) with a factor of $1/t$ outside. $\endgroup$ Commented Apr 30, 2013 at 16:57

This answer applies to the question in its form on Wed May 14 2014. It seems to have escaped attention that this sum may be evaluated using harmonic summation techniques. This will provide another integral representation by a complex line integral.

To see this, introduce the sum $$S(x) = \sum_{k\ge 1} \left(\frac{1}{(2k-1)x}\right)^2 \frac{1}{1-i((2k-1)x)^2}$$ with $x$ a real number.

This sum converges absolutely and it is harmonic and may be evaluated by inverting its Mellin transform.

Recall the harmonic sum identity $$\mathfrak{M}\left(\sum_{k\ge 1} \lambda_k g(\mu_k x);s\right) = \left(\sum_{k\ge 1} \frac{\lambda_k}{\mu_k^s} \right) g^*(s)$$ where $g^*(s)$ is the Mellin transform of $g(x).$

In the present case we have $$\lambda_k = 1, \quad \mu_k = 2k-1 \quad \text{and} \quad g(x) = \frac{1}{x^2}\frac{1}{1-ix^2}.$$

We need the Mellin transform $g^*(s)$ of $g(x)$ which is $$\int_0^\infty \frac{1}{x^2}\frac{1}{1-ix^2} \times x^{s-1} dx = \int_0^\infty \frac{1}{1-ix^2} \times x^{s-3} dx.$$

Compute the fundamental strip of this transform. In a neighborhood of zero, $g(x) \sim 1/x^2$ and we need $(s-1)-2 > -1$ so the integral converges when $\Re(s)>2$. At infinity $g(x) \sim 1/x^4$ which gives convergence for $(s-1)-4<-1$ or $\Re(s) < 4.$ Therefore the fundamental strip is $\langle 2, 4 \rangle.$

We evaluate $g^*(s)$ by using a semicircular contour in the upper half plane with radius $R$ going to infinity traversed counterclockwise. Along the positive real axis we get the transform $g^*(s).$ Along the negative real axis we get

$$\int_{-\infty}^0 \frac{1}{1-ix^2} \times x^{s-3} dx = - \int_\infty^0 \frac{1}{1-iu^2} \times e^{i\pi (s-3)} u^{s-3} du \\ = e^{i\pi (s-1)} g^*(s) = - e^{i\pi s} g^*(s).$$

The sole pole in the upper half plane is at $x=e^{3/4\pi i}$ and since the pole is simple we have $$\mathrm{Res}\left(\frac{1}{1-ix^2} x^{s-3}; x=e^{3/4\pi i} \right) = \frac{1}{-2i\times e^{3/4\pi i}} e^{3/4\pi i(s-3)} \\ = \frac{i}{2} e^{3/4\pi i(s-4)} = -\frac{i}{2} e^{3/4\pi i s}.$$ This yields for $g^*(s)$ that $$g^*(s) (1 - e^{i\pi s}) = 2\pi i \times -\frac{i}{2} e^{3/4\pi i s}$$ so that $$g^*(s) = \pi \frac{ e^{3/4\pi i s} }{1 - e^{i\pi s}} = \pi e^{1/4\pi i s} \frac{1}{e^{-1/2 i\pi s} - e^{1/2 i\pi s}} \\= - \frac{\pi}{2i} \frac{e^{1/4\pi i s}}{\sin(1/2 \pi s)} = \frac{\pi i}{2} \frac{e^{1/4\pi i s}}{\sin(1/2 \pi s)}.$$

It follows that the Mellin transform $Q(s)$ of the harmonic sum $S(x)$ is given by

$$Q(s) = \frac{\pi i}{2} \frac{e^{1/4\pi i s}}{\sin(1/2 \pi s)} \left(1-\frac{1}{2^s}\right)\zeta(s) \\ \text{because}\quad \sum_{k\ge 1} \frac{\lambda_k}{\mu_k^s} = \sum_{k\ge 1} \frac{1}{(2k-1)^s} = \left(1-\frac{1}{2^s}\right)\zeta(s)$$ for $\Re(s) > 1.$

The Mellin inversion integral here is $$\frac{1}{2\pi i} \int_{5/2-i\infty}^{5/2+i\infty} Q(s)/x^s ds$$ which we evaluate by shifting it to the right for an expansion about infinity. The choice of line was determined by the fundamental strip of $g^*(s)$ which is completely embedded in the half-plane of convergence $\Re(s)>1$ of the sum term.

We now proceed to collect the contribution in residues from the poles. These originate in the sine term only at values $s=2q$ with $q\ge 2,$ giving $$\frac{\pi i}{2} \sum_{q\ge 2} e^{1/2\pi i q} \times (-1)^q \frac{2}{\pi} \times \left(1-\frac{1}{2^{2q}}\right) \zeta(2q) \times \frac{1}{x^{2q}} \\ = i \sum_{q\ge 2} e^{3/2\pi i q} \times \left(1-\frac{1}{2^{2q}}\right) \zeta(2q) \times \frac{1}{x^{2q}}.$$ To simplify this we need to evaluate the sum $$T(u) = \sum_{q\ge 2} u^{2q} \zeta(2q) = \sum_{q\ge 2} u^{2q} \frac{(-1)^{q+1} B_{2q} (2\pi)^{2q}}{2(2q)!} = - \frac{1}{2} \sum_{q\ge 2} B_{2q} \frac{(2\pi i u)^{2q}}{(2q)!}.$$

Using the generating function of the Bernoulli numbers we have that $$\sum_{p\ge 4} B_p \frac{t^p}{p!} = -1 + \frac{1}{2} t - \frac{1}{12} t + \frac{t}{e^t-1}$$ which gives for the sum $$\frac{1}{2} - \frac{1}{4}(2\pi i u) + \frac{1}{24} (2\pi i u)^2 - \frac{1}{2} \frac{2\pi i u}{e^{2\pi i u} - 1}.$$ The expansion that we have is given by $$i T(e^{3/4 \pi i}/x) - i T(e^{3/4 \pi i}/x/2)$$

The closed form for the sum is actually the negative of this value because we are integrating counterclockwise. Doing the substitution we finally obtain $${\frac {1/8\,\sqrt {2}\pi -1/8\,i\pi \,\sqrt {2}}{x}} +1/8\,{\frac {{\pi }^{2}}{{x} ^{2}}} - \frac{\pi e^{3/4 \pi i}/x}{e^{2\pi i e^{3/4 \pi i}/x}-1} + \frac{\pi e^{3/4 \pi i}/x/2}{e^{2\pi i e^{3/4 \pi i}/x/2}-1}.$$ This expansion holds everywhere the original series converges, i.e. $(-\infty, 0)\cup (0, \infty).$


You can have the following closed form for the sum

$$ -\frac{\sqrt{2}\left( 1+\,i \right) \pi}{32\,t^2} \, \left( (i-1)\pi \,\sqrt {2}+4\,t\,\tan \left( {\frac { \left( 1-i \right) \pi \,\sqrt {2}}{4\,t}} \right) \right). $$

Check Here is the original answer by Maple in case I made a mistake while trying to improve the format in the above answer

$$ \left( -1/32-1/32\,i \right) \pi \, \left( i\pi \,\sqrt {2}+4\,\tan \left( {\frac { \left( 1/4-1/4\,i \right) \pi \,\sqrt {2}}{t}} \right) t-\pi \,\sqrt {2} \right) \sqrt {2}{t}^{-2}.$$

  • $\begingroup$ Thanks, I got the same answer with Mathematica too. I'm really looking for an analytical way to find the sum. Even a good approximation for $S$, or an approximation valid under certain restrictions on $t$ would work just fine. $\endgroup$ Commented Apr 30, 2013 at 19:05
  • 1
    $\begingroup$ @G.H.Hardly: It is a nice closed form in terms of elementary functions! $\endgroup$ Commented May 4, 2013 at 2:38

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