Evaluating a class of infinite sums in closed form

Another way to get to the same result is to use &quot;Feynman&#x27;s Trick&quot; of differentiating inside a sum:<p>Consider the function f(x) = Sum_{n=1}^\infty c^(-xn)<p>Then differentiate this k times. Each time you pull down a factor of n (as well as a log(c), but that&#x27;s just a constant). So, the sum you&#x27;re looking for is related to the kth derivative of this function.<p>Now, fortunately this function can be evaluated explicitly since it&#x27;s just a geometric series: it&#x27;s 1 &#x2F; (c^x - 1) -- note that the sum starts at 1 and not 0. Then it&#x27;s just a matter of calculating a bunch of derivatives of this function, keeping track of factors of log(c) etc. and then evaluating it at x = 1 at the very end. Very labor intensive, but (in my opinion) less mysterious than the approach shown here (although, of course the polylogarithm function is precisely this tower of derivatives for negative integer values).

This is pretty neat! I was toying around with the problem and it appears you can use generating functions to derive the same sequence of operations. If you start with:<p><pre><code> G(x) = 1 + x + x^2 + ... = 1&#x2F;(1-x) </code></pre> The coefficients of this polynomial is the sequence (0^0, 1^0, 2^0, ...)<p>If you take the derivative of G(x) and multiply by x you get:<p><pre><code> x * G&#x27;(x) = x + 2*x^2 + 3*x^3 + ... = x * d&#x2F;dx 1&#x2F;(1-x) = x&#x2F;(1-x)^2 </code></pre> The coefficients of this polynomial is the sequence (0^1, 1^1, 2^1, ...). If you repeat this step, you get a polynomial whose coefficients are (0^2, 1^2, 2^2, ...) and if you do this operation N times, you can get a closed form of a polynomial whose coefficients are (0^N, 1^N, 2^N, ...).<p>The infinite sum converges for -1 &lt; x &lt; 1. If you set x=1&#x2F;c, you get the infinite sum<p><pre><code> 0^N&#x2F;c^0 + 1^N&#x2F;c^1 + 2^N&#x2F;c^2 + ... </code></pre> which is exactly the sum we are trying to solve for. This means you solve any infinite sum of the form given by taking the derivative of 1&#x2F;(1-x) N times while multiplying by x each time. Then plug in x=1&#x2F;c at the end.

Aside: it looks like the c=2 case generates only integers, and it&#x27;s an OEIS sequence which shows up in a lot of combinatorics problems,<p><a href="https:&#x2F;&#x2F;oeis.org&#x2F;A076726" rel="nofollow">https:&#x2F;&#x2F;oeis.org&#x2F;A076726</a> (<i>&quot; A076726 | a(n) = Sum_{k&gt;=0} k^n&#x2F;2^k&quot;</i>)<p><a href="https:&#x2F;&#x2F;oeis.org&#x2F;A000629" rel="nofollow">https:&#x2F;&#x2F;oeis.org&#x2F;A000629</a> (<i>&quot;A000629 | Number of necklaces of partitions of n+1 labeled beads&quot;</i>)

I found it useful to walk through evaluation of a few elementary instances of this class using simpler methods, to put the main result in perspective. Specifically, replace the initial 3 exponent with 0 or 1.<p>If the exponent is 0, then you have the sum 1&#x2F;2 + 1&#x2F;4 + 1&#x2F;8 + 1&#x2F;16 + 1&#x2F;32 + ..., from Zeno&#x27;s most famous paradox (<a href="https:&#x2F;&#x2F;en.wikipedia.org&#x2F;wiki&#x2F;Zeno%27s_paradoxes" rel="nofollow">https:&#x2F;&#x2F;en.wikipedia.org&#x2F;wiki&#x2F;Zeno%27s_paradoxes</a> ). If you are fortunate, you previously learned that this converges to 1, and played around with this enough in your head to have a solid understanding of why. If you are less fortunate, I recommend pausing to digest this result.<p>Then, if the exponent is 1, you have the sum 1&#x2F;2 + 2&#x2F;4 + 3&#x2F;8 + 4&#x2F;16 + 5&#x2F;32 + ... .<p>What happens if we subtract (1&#x2F;2 + 1&#x2F;4 + 1&#x2F;8 + 1&#x2F;16 + 1&#x2F;32 + ...) from it? We have (1&#x2F;4 + 2&#x2F;8 + 3&#x2F;16 + 4&#x2F;32 + ...) left over.<p>Then, if we subtract (1&#x2F;4 + 1&#x2F;8 + 1&#x2F;16 + 1&#x2F;32 + ...) from the latter, we still have (1&#x2F;8 + 2&#x2F;16 + 3&#x2F;32 + ...) left over.<p>Then, if we subtract (1&#x2F;8 + 1&#x2F;16 + 1&#x2F;32 + ...) from the latter, we still have (1&#x2F;16 + 2&#x2F;32 + ...) left over.<p>Continuing in this fashion, we end up subtracting off<p>(1&#x2F;2 + 1&#x2F;4 + 1&#x2F;8 + 1&#x2F;16 + 1&#x2F;32 + ...) + (1&#x2F;4 + 1&#x2F;8 + 1&#x2F;16 + 1&#x2F;32 + ...) + (1&#x2F;8 + 1&#x2F;16 + 1&#x2F;32 + ...) + (1&#x2F;16 + 1&#x2F;32 + ...) + (1&#x2F;32 + ...) + ...<p>and this converges to the main sum. And, from the exponent-0 result, we know this is just 1 + 1&#x2F;2 + 1&#x2F;4 + 1&#x2F;8 + 1&#x2F;16 + ...

Hi in the 3rd equation you meant to write Li_s(z) but you wrote x instead.<p>That was an interesting article thanks.

Fun stuff. A good follow-up is the book: A = B, by Marko Petkovsek, Herbert S Wilf, Doron Zeilberger, A K Peters&#x2F;CRC Press, 1996.

It feels odd the sum of the first example is 26. (All the number nerds, please forgive me). It&#x27;s such a usual number.

The answer is just 26? That’s crazy.<p>Wonder if there’s any way to intuit this before working it out.

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Off-topic question: I’m using the iOS browser extension Noir, which adds dark-mode support to web sites that don’t support dark mode by themselves. However, this causes MathJax(?) formulas like in the article to be displayed black on black. Does anyone know of a similar browser extension that can handle this? (And yes, I already reported this issue to Noir some time ago.)

the title makes this seem like some major or original discovery in math. Try evaluating the logarithmic integral with positive n , like r^3&#x2F;n^3 instead of n^3&#x2F;r^3. Way harder and more interesting.