Euler Sums, Part I

Sums of the form \displaystyle E(p,q) = \sum_{n=1}^{\infty} \frac{H^{(p)}_n}{n^q}, where \displaystyle H^{(p)}_n = \sum_{k=1}^n \frac{1}{n^p} (i.e., the nth harmonic number of order p) are sometimes called Euler sums.  There are a large number of interesting ways to evaluate these Euler sums, such as converting them to integrals, applying complex analysis, and rewriting them as polylogarithms.  I tend to prefer the most basic techniques, though – those that involve manipulating the sum directly.

This post reproduces one of those basic techniques for evaluating \displaystyle \sum_{n=1}^{\infty} \frac{H_n}{n^2}.  It appeared as part of an answer by Rob Johnson to a question of mine about the evaluation of three different alternating Euler sums.  (An alternating Euler sum is one of the form \displaystyle \sum_{n=1}^{\infty} \frac{(-1)^{n-1} H^{(p)}_n}{n^q}.)

Here’s Rob’s argument.

\displaystyle \sum_{n=1}^{\infty} \frac{H_n}{n^2} = \sum_{n=1}^{\infty} \sum_{k=1}^{\infty} \frac{1}{n^2} \left( \frac{1}{k} - \frac{1}{k+n} \right)

\displaystyle = \sum_{n=1}^{\infty} \sum_{k=1}^{\infty} \frac{1}{n k (k+n)}

\displaystyle = \sum_{k=1}^{\infty} \sum_{n=k+1}^{\infty} \frac{1}{(n-k)kn}

\displaystyle = \sum_{n=2}^{\infty} \sum_{k=1}^{n-1} \frac{1}{(n-k)kn}

\displaystyle = \sum_{n=2}^{\infty} \sum_{k=1}^{n-1} \frac{1}{n^2} \left(\frac{1}{k} + \frac{1}{n-k}\right)

\displaystyle = 2\sum_{n=1}^{\infty} \frac{1}{n^2} H_{n-1}

\displaystyle = 2\sum_{n=1}^{\infty} \frac{1}{n^2} H_n - 2\sum_{n=1}^{\infty} \frac{1}{n^3}

\displaystyle \implies \sum_{n=1}^{\infty} \frac{H_n}{n^2} = 2 \zeta(3).

This is a clever sequence of manipulations.  The first good idea is to express H_n as a telescoping series in the first step.  The variable switch in step three (n-k for n) is nice because, after swapping the order of summation in step four and the partial fractions decomposition in step five, it sets up the crucial step six: \displaystyle \sum_{k=1}^{n-1} \left(\frac{1}{k} + \frac{1}{n-k}\right) is just H_{n-1} added forwards and backwards (and so twice).  (The switch in the lower index in step six from n = 2 to n = 1 is because H_0 is an empty sum and so is 0.)  Step seven is also nice: Adding two copies of \displaystyle \sum_{n=1}^{\infty} \frac{1}{n^3} to the sum in line six enables you to change the H_{n-1} to H_n, giving you two copies of the Euler sum you’re trying to evaluate on the right side of the equation in addition to the one already one the left side.  Of course, if you add two copies of \displaystyle \sum_{n=1}^{\infty} \frac{1}{n^3} you have to subtract two copies as well, and the basic algebra that remains in step eight gives you the final, simple, beautiful, result:

\displaystyle \sum_{n=1}^{\infty} \frac{H_n}{n^2} = 2\zeta(3).

For more information on evaluating Euler sums, you can see the rest of Rob’s answer or some of the other nice answers to my question about evaluating some simple alternating Euler sums.  (The integral representations in this answer are particularly nice, but the others are good, too.)  Michael Hoffman’s page on multiple zeta values and Euler sums contains a huge number of references for evaluation of Euler sums and their variations.  It’s a great resource.  I’ve also recently been thumbing through Ovidiu Furdui’s problem book Limits, Series, and Fractional Part Integrals [1].  Chapter 3, “A Bouquet of Series,” contains explicit evaluation of several Euler and Euler-type sums.

References

1.  Ovidiu Furdui, Limits, Series, and Fractional Part Integrals: Problems in Mathematical Analysis, Springer Problem Books in Mathematics, 2013.

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This entry was posted in calculus, euler sums, real analysis, sequences and series. Bookmark the permalink.

One Response to Euler Sums, Part I

  1. Pingback: Euler Sums, Part II: A Symmetry Formula | A Narrow Margin

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