Huygens’ problem to Leibnitz: solution

In the Feb 23 2018 blog problem, we posed the following question:

Sum the following infinite series:

$1+\frac{1}{3} + \frac{1}{6} + \frac{1}{10} + \frac{1}{15}+ \ldots$ .

Solution:

The sum can be written as:

$S=\sum_{n=1}^{\infty}P_{n}$ , where $P_{n}=\frac{2}{n(n+1)}=2(\frac{1}{n}-\frac{1}{n+1})$ .

Thus, $2(1-\frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + \ldots)=2$ . This is the answer.

If you think deeper, this needs some discussion about rearrangements of infinite series also. For the time, we consider it outside our scope.

Cheers,

Nalin Pithwa.

Enter your comment here...

Fill in your details below or click an icon to log in:

Email (required) (Address never made public)

Name (required)

Website

You are commenting using your WordPress.com account. ( Log Out / Change )

You are commenting using your Google account. ( Log Out / Change )

You are commenting using your Twitter account. ( Log Out / Change )

You are commenting using your Facebook account. ( Log Out / Change )

Connecting to %s

%d bloggers like this: