0
$\begingroup$

I asked a question about this series;

$(1 - \frac12)+(\frac13 - \frac14)(1 - \frac12 + \frac13)+(\frac15 - \frac16)(1 - \frac12 + \frac13 - \frac14 + \frac15)+(\frac17 - \frac18)(1 - \frac12 + \frac13 - \frac14 + \frac15 - \frac16 + \frac17)+...$

in a previous thread and something else about it that I'd like to know is if there is a name for series where the coefficient of each term is a partial sum? Furthermore, is there a general method for finding the closed form sums of such series?

  • 0
    Some keywords for such kind of sums of sums [of sums] : [Euler or Euler-Zagier sums](http://mathworld.wolfram.com/EulerSum.html), [Multiple Zeta Value](http://en.wikipedia.org/wiki/Multiple_zeta_function) with many papers with the name [Broadhurst](http://arxiv.org/abs/0907.2557v2) Borwein and so on, see too [Multiple Polylogarithms](http://arxiv.org/abs/math.CA/0310062) (in the last one you'll see that alternate sums are considered too).2012-12-28

1 Answers 1

3

There is no special name since what you have is just a double summation instead of a single summation. Your series is nothing but $\sum_{n=0}^{\infty} \sum_{k=1}^{2n+1} \left(\dfrac1{2n+1}-\dfrac1{2n+2}\right) \left(\dfrac{(-1)^{k-1}}{k} \right)$ which can also be written as a single summation $\sum_{n=0}^{\infty} \left(\dfrac1{2n+1}-\dfrac1{2n+2}\right) \left(H_{2n+1} - H_n\right)$

  • 0
    Ah so it is a double sum. Thought so. Sorry to bother you with such a trivial question it was just that sometimes it's written with the sigma in the middle which somehow confused me!2012-12-29