How can the exact result of the summation with a+nb be calculated?

[arpon]

Homework Statement

$$ \sum_{n=1}^\infty\frac{1}{1+(a+nb)^2} = ? $$

2. The attempt at a solution
I approximated the result by integration,

$$
\begin{align}
\sum_{n=1}^\infty \frac{1}{1+(a+nb)^2} &\approx \lim_{N \rightarrow +\infty} {\int_{0}^N \frac{1}{1+(a+bx)^2} dx}\\
&= \lim_{N \rightarrow +\infty} {\frac{tan^{-1} (a + Nb)}{b}} - \frac{tan^{-1} (a)}{b}\\
&= \frac{\pi}{2b} - \frac{tan^{-1} (a)}{b}
\end{align}
$$
Using this integration method, I further proved,
$$\frac{\pi}{2b} - \frac{tan^{-1} (a+b)}{b}< \sum_{n=1}^\infty \frac{1}{1+(a+nb)^2} < \frac{\pi}{2b} - \frac{tan^{-1} (a)}{b}$$

But, how can I calculate the exact result?

I faced this problem when solving a physics problem.

 

[STEMucator]

I believe you are trying to use a theorem that you probably should have mentioned in the relevant equations. It may have been helpful for you to list it.

For anyone who might be interested in what the theorem says, suppose $a_n = f(n)$ where the graph of $f(x)$ is positive on $[1, \infty)$. Suppose further $f(x)$ is decreasing and concave up, and the improper integral:

$$\int_1^{\infty} f(x) dx$$

converges. Then:

$$S_n + \int_{n+1}^{\infty} f(x) \space dx + \frac{a_{n+1}}{2} < S < S_n + \int_{n}^{\infty} f(x) \space dx - \frac{a_{n+1}}{2}$$

Where $S$ is the sum of the series, and $S_n$ is a partial sum of the series.

The bound you obtain on the sum is relatively tight. A better approximation of $S$ might be to take the two results and average them.

Now, the value of $n$ you choose to use is important. If you want a small amount of error in your estimate, try choosing $n$ such that:

$$\int_{n}^{\infty} f(x) \space dx - \frac{a_{n+1}}{2} - \left[ \int_{n+1}^{\infty} f(x) \space dx + \frac{a_{n+1}}{2} \right] < 0.00001$$

Where the quantity on the left hand side of the inequality is the width of the interval. This will give you an estimate of the form $p < S < q$ where you can say with a good deal of certainty $S ≈ \frac{p + q}{2}$.

 

[Ray Vickson]

Homework Statement

$$ \sum_{n=1}^\infty\frac{1}{1+(a+nb)^2} = ? $$

2. The attempt at a solution
I approximated the result by integration,

$$
\begin{align}
\sum_{n=1}^\infty \frac{1}{1+(a+nb)^2} &\approx \lim_{N \rightarrow +\infty} {\int_{0}^N \frac{1}{1+(a+bx)^2} dx}\\
&= \lim_{N \rightarrow +\infty} {\frac{tan^{-1} (a + Nb)}{b}} - \frac{tan^{-1} (a)}{b}\\
&= \frac{\pi}{2b} - \frac{tan^{-1} (a)}{b}
\end{align}
$$
Using this integration method, I further proved,
$$\frac{\pi}{2b} - \frac{tan^{-1} (a+b)}{b}< \sum_{n=1}^\infty \frac{1}{1+(a+nb)^2} < \frac{\pi}{2b} - \frac{tan^{-1} (a)}{b}$$

But, how can I calculate the exact result?

I faced this problem when solving a physics problem.

I doubt that the summation has a nice closed-form in terms of elementary functions, but it is expressible in terms of the non-elementary "di-Gamma" function. Maple gets
\text{answer} = \frac{i}{2b} \left( \Psi \left( 1 + \frac{a-i}{b} \right) - \Psi \left( 1 + \frac{a+i}{b} \right) \right),
where $\Psi$ is the so-called di-Gamma function, defined as
\Psi(x) = \frac{d}{dx} \ln (\Gamma (x)) = \frac{1}{\Gamma (x)} \cdot \frac{d}{dx} \Gamma (x)
and $i = \sqrt{-1}$. Despite its complex appearance, the result is real if $a,b$ are real. Here is a Maple plot for $0 \leq a \leq 1$ and $0 \leq b \leq 10$. (I accidentally made the plot with different upper bounds on $a$ and $b$; I had intended to go from 0 to 1 for both. However, once I had uploaded the unintended plot the PF editor would not let me delete it and replace it by the correct one, without deleting the entire post. Anyway, the plot with $0 \leq b \leq 1$ does not look much different from the one attached.)

 

[Ray Vickson]

Homework Statement

$$ \sum_{n=1}^\infty\frac{1}{1+(a+nb)^2} = ? $$

2. The attempt at a solution
I approximated the result by integration,

$$
\begin{align}
\sum_{n=1}^\infty \frac{1}{1+(a+nb)^2} &\approx \lim_{N \rightarrow +\infty} {\int_{0}^N \frac{1}{1+(a+bx)^2} dx}\\
&= \lim_{N \rightarrow +\infty} {\frac{tan^{-1} (a + Nb)}{b}} - \frac{tan^{-1} (a)}{b}\\
&= \frac{\pi}{2b} - \frac{tan^{-1} (a)}{b}
\end{align}
$$
Using this integration method, I further proved,
$$\frac{\pi}{2b} - \frac{tan^{-1} (a+b)}{b}< \sum_{n=1}^\infty \frac{1}{1+(a+nb)^2} < \frac{\pi}{2b} - \frac{tan^{-1} (a)}{b}$$

But, how can I calculate the exact result?

I faced this problem when solving a physics problem.

You can use the Euler-Maclaurin sum formula to get more accurate estimates of $\sum_{n=1}^N 1/[1+(a + bn)^2]$ for finite, large $N$. See, eg.,

https://en.wikipedia.org/wiki/Euler–Maclaurin_formula or
http://people.csail.mit.edu/kuat/courses/euler-maclaurin.pdf (for infinite sum, too).

Some of these approximations may be much better than the one you used above. In any case, they allow you to bound the error in that approximation.