Threaded Mode | Linear Mode

Thomas Klemm · (This post was last modified: 07-26-2022 01:47 PM by Thomas Klemm.)

(07-23-2022 02:20 AM)Thomas Klemm Wrote: It uses Ramanujan's formula:

\(
\begin{align}
\pi^2 = 10 - \sum_{n=1}^{\infty} \frac{1}{n^3(n+1)^3}
\end{align}
\)

As I wondered why this should be the case I came up with the following derivation:

\(
\begin{align}
\frac{1}{n} - \frac{1}{n+1}
&= \frac{n+1 - n}{n(n+1)} \\
&= \frac{1}{n(n+1)} \\
\end{align}
\)

Plug this into the infinite sum to get:

\(
\begin{align}
\sum_{n=1}^{\infty} \frac{1}{n^3(n+1)^3}
&= \sum_{n=1}^{\infty} \frac{1}{[n(n+1)]^3} \\
&= \sum_{n=1}^{\infty} \left[\frac{1}{n} - \frac{1}{n+1}\right]^3 \\
\end{align}
\)

That's a bit like a telescoping sum.
It is of the form:

\(
\begin{align}
[a-b]^3 + [b-c]^3 + [c-d]^3 + \cdots
&= a^3 - 3a^2b + 3ab^2 - b^3 \\
&+ b^3 - 3b^2c + 3bc^2 - c^3 \\
&+ c^3 - 3c^2d + 3cd^2 - d^3 \\
&+ \cdots
\end{align}
\)

We notice that \(b^3\), \(c^3\), \(d^3\), \(\cdots\) all cancel.
Also we can extract the common factors of the residual terms:

\(
\begin{align}
[a-b]^3 + [b-c]^3 + [c-d]^3 + \cdots
= a^3 &- 3ab(a-b) \\
&- 3bc(b-c) \\
&- 3cd(c-d) \\
&+ \cdots
\end{align}
\)

This leaves us with:

\(
\begin{align}
\sum_{n=1}^{\infty} \frac{1}{n^3(n+1)^3}
&= 1 - 3 \sum_{n=1}^{\infty} \frac{1}{n} \frac{1}{n+1} \left[\frac{1}{n} - \frac{1}{n+1}\right] \\
&= 1 - 3 \sum_{n=1}^{\infty} \frac{1}{n} \frac{1}{n+1}\frac{1}{n(n+1)} \\
&= 1 - 3 \sum_{n=1}^{\infty} \frac{1}{[n(n+1)]^2} \\
\end{align}
\)

A quick sanity check shows: Sum[1/(n(n+1))^2,{n,1,∞}]

\(
\begin{align}
\sum_{n=1}^{\infty} \frac{1}{[n(n+1)]^2} = \frac{1}{3}\left(\pi^2-9\right)
\end{align}
\)

It appears that we're on track.

We can do the same as before but now the squares don't cancel:

\(
\begin{align}
[a-b]^2 + [b-c]^2 + [c-d]^2 + \cdots
&= a^2 - 2ab + b^2 \\
&+ b^2 - 2bc + c^2 \\
&+ c^2 - 2cd + d^2 \\
&+ \cdots
\end{align}
\)

Instead they are added twice (except for the first) and so we get:

\(
\begin{align}
\sum_{n=1}^{\infty} \frac{1}{[n(n+1)]^2}
&= 2 \sum_{n=1}^{\infty} \frac{1}{n^2} - 1 \\
&- 2 \sum_{n=1}^{\infty} \frac{1}{n(n+1)} \\
\end{align}
\)

But the second sum is a telescoping sum:

\(
\begin{align}
\sum_{n=1}^{\infty} \frac{1}{n(n+1)}
&= \sum_{n=1}^{\infty} \frac{1}{n} - \frac{1}{n+1} \\
&= 1
\end{align}
\)

And thus we end up with:

\(
\begin{align}
\sum_{n=1}^{\infty} \frac{1}{[n(n+1)]^2}
&= 2 \sum_{n=1}^{\infty} \frac{1}{n^2} - 3 \\
\end{align}
\)

But now we're in well known territory since:

\(
\begin{align}
\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6} \\
\end{align}
\)

This leads to:

\(
\begin{align}
\sum_{n=1}^{\infty} \frac{1}{[n(n+1)]^2}
&= 2 \frac{\pi^2}{6} - 3 \\
&= \frac{\pi^2}{3} - 3 \\
&= \frac{1}{3}\left(\pi^2-9\right) \\
\end{align}
\)

Plug this into the formula above and we get:

\(
\begin{align}
\sum_{n=1}^{\infty} \frac{1}{n^3(n+1)^3}
&= 1 - 3 \sum_{n=1}^{\infty} \frac{1}{[n(n+1)]^2} \\
&= 1 - 3 \frac{1}{3}\left(\pi^2-9\right) \\
&= 1 - \pi^2 + 9 \\
&= 10 - \pi^2 \\
\end{align}
\)

References