1

I am trying to approximate $\sum \frac{1}{n^2}$ with telescopic series. So far, I've understood up to this: $$\begin{eqnarray*} \sum_{n\geq m}\frac{1}{n^2}&=&\sum_{n\geq m}\left(\frac{1}{n}-\frac{1}{(n+1)}\right)+\frac{1}{2}\sum_{n\geq m}\left(\frac{1}{n^2}-\frac{1}{(n+1)^2}\right)\\&+&\frac{1}{6}\sum_{n\geq m}\left(\frac{1}{n^3}-\frac{1}{(n+1)^3}\right)-\frac{1}{6}\sum_{n\geq m}\frac{1}{n^3(n+1)^3}\end{eqnarray*} $$ I am told that the last term, $-\displaystyle\frac{1}{6}\sum_{n\geq m}\frac{1}{n^3(n+1)^3}$, can be represented as $C\sum(\frac{C}{n^5}-\frac{C}{(n+1)^5})$ plus sum other higher order term, where $C$ is a just a number, different in each appearance.

Questions

How do we make this cubic into a telescopic sum of quintics?

And after we do that, how do we keep going and reating more and more telescopic terms, like $C\sum(\frac{C}{n^7}-\frac{C}{(n+1)^7})$ or something?

D.R.
  • 8,691
  • 4
  • 22
  • 52
  • What is the asymptotic behaviour of $\frac{1}{n^3(n+1)^3}$ as $n\to +\infty$? What is the asymptotic behaviour of $\frac{C}{n^5}-\frac{C}{(n+1)^5}$? For which value of $C$ the leading term of the asymptotic expansion is the same? Pretty straightforward computations. – Jack D'Aurizio Aug 23 '17 at 04:59
  • And by the Euler-MacLaurin summation formula, the coefficients you get by such approach are clearly related with Bernoulli numbers. – Jack D'Aurizio Aug 23 '17 at 05:00

1 Answers1

1

$$\frac{1}{n^3(n+1)^3}=\frac{1}{n^6}-\frac{3}{n^7}+\frac{6}{n^8}+\ldots \tag{1}$$ $$\frac{1}{n^5}-\frac{1}{(n+1)^5} = \frac{5}{n^6}-\frac{15}{n^7}+\frac{35}{n^8}\ldots\tag{2} $$ hence $$ \frac{1}{n^3(n+1)^3}-\color{blue}{\frac{1}{5}}\left(\frac{1}{n^5}-\frac{1}{(n+1)^5}\right) = -\frac{1}{n^8}+\ldots \tag{3} $$ where the LHS equals $$ -\frac{1+5 n+5 n^2}{5 n^5 (1+n)^5} \tag{4}$$ which can be approximated by something of the form $\frac{D}{n^7}-\frac{D}{(n+1)^7}$
by just repeating the steps $(1),(2),(3)$.

I am quite fond of the celebrity my proof of Stirling's approximation is acquiring. I got that idea while writing the first section of these notes, before discovering such approach was already well-known in the literature.

Jack D'Aurizio
  • 353,855
  • How do we find the asymptotic approximation $\frac{1}{n^6} - \frac{3}{n^7} + \ldots$? – D.R. Aug 25 '17 at 05:01
  • @D.R.: by computing a Taylor series. – Jack D'Aurizio Aug 25 '17 at 05:36
  • Is it a coincidence that it skips $n^4$ and $n^6$, or is there an underlying reason? – D.R. Sep 03 '17 at 15:04
  • @D.R.: not a coincidence at all, that depends on the fact that Bernoulli numbers with odd index are always zero, with the only exception of $B_1$. – Jack D'Aurizio Sep 03 '17 at 18:58
  • Ok. I now understand how we got to $\frac{1}{z}+\frac{1}{2z^2}+\frac{1}{6z^3}-\frac{1}{30z^5}\leq\psi'\leq\frac{1}{z}+\frac{1}{2z^2}+\frac{1}{6z^3}$, but how do I proceed from there? Do I integrate all 3 sides, or do I approximate $\psi' \approx \frac{1}{z}+\frac{1}{2z^2}+\frac{1}{6z^3}$ and integrate that? – D.R. Sep 04 '17 at 04:29