Series summation of rational function with quadratic denominator

The goal of this page is to consider infinite series summations of the form:

$\sum _{k=k_0}^{\mathrm{\infty }}\frac{1}{\text{quadratic function of }k}$

where the denominator function is nonzero for all summands.

Our overall goal is as follows:

• Identify some situations where there exist closed-form expressions for the solutions, and find the solutions in those cases.
• For the remaining situations, find good upper and lower bounds on the infinite summation.

Case of denominator having linear factors that differ by a nonzero integer

In this case, we can use additive telescoping to get an explicit expression for the series sum. Explicitly, consider a summation:

$\sum _{k=k_0}^{\mathrm{\infty }}\frac{1}{(k-\alpha )(k-(\alpha -n))}$

Each summand can be written as:

$\frac{1}{(k-\alpha )(k-(\alpha -n))}=\frac{1}{n}(\frac{1}{k-\alpha }-\frac{1}{k-(\alpha -n)})$

We can now telescope the partial sums and take the limit. The final answer is:

$\sum _{k=k_0}^{\mathrm{\infty }}\frac{1}{(k-\alpha )(k-(\alpha -n))}=\frac{1}{n}(\sum _{m=0}^{n-1}\frac{1}{k_0-(\alpha -m)})$

Note that we have converted an infinite sum problem to adding up a finite number of fractions. With explicit numerical values of $k_0,\alpha ,m$, we can calculate the answer explicitly.

Case of denominator a perfect square

Here, we are considering a summation of the form:

$\sum _{k=k_0}^{\mathrm{\infty }}\frac{1}{(k-\alpha {)}^2}$

Doing this in general is hard. However, there is a useful fact:

$\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k^2}=\frac{{\pi }^2}{6}$

Using this fact, we can tackle the case $\alpha =0$, and more generally, any situation where $\alpha$ is an integer. A little manipulation of the fact gives us that:

$\sum _{k=0}^{\mathrm{\infty }}\frac{1}{(2k+1{)}^2}=\frac{{\pi }^2}{8}$

This allows us to tackle the case where $\alpha$ is half of an odd integer.

Case of denominator having linear factors that differ by half an odd integer

Not all these cases can be handled, but some can, using a clever trick.

Consider:

$\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k(k-(1/2))}$

The terms can be written as:

$\frac{1}{k(k-(1/2))}=2(\frac{1}{k-(1/2)}-\frac{1}{k})=4(\frac{1}{2k-1}-\frac{1}{2k})$

The summation explicitly is:

$4(1-\frac{1}{2})+4(\frac{1}{3}-\frac{1}{4})+4(\frac{1}{5}-\frac{1}{6})+\dots$

If we rearrange parentheses a bit (without affecting the overall order of addition), we get:

$4[1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\dots ]$

The inner summation is well known for being $\ln 2$, so we get:

$\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k(k-(1/2))}=4\ln 2$

Note that although the inner summation (after we've dropped the parentheses) is only conditionally convergent, the original summation is something that we know is absolutely convergent.