Series summation of rational function with quadratic denominator

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

$\sum _{k=k_{0}}^{\infty }{\frac {1}{{\mbox{quadratic function of }}k}}$

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:

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

Each summand can be written as:

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

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

$\sum _{k=k_{0}}^{\infty }{\frac {1}{(k-\alpha )(k-(\alpha -n))}}={\frac {1}{n}}\left(\sum _{m=0}^{n-1}{\frac {1}{k_{0}-(\alpha -m)}}\right)$

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:

$\sum _{k=k_{0}}^{\infty }{\frac {1}{(k-\alpha )^{2}}}$

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

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

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:

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

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:

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

The terms can be written as:

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

The summation explicitly is:

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

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

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

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

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

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.