Solution To Math Problem Posed To iBot

Hello iBot and CH,

iBot requested that I show how a problem I posed to it is solved: iBot's Request

The problem can be found here, and I am posting the solution here so I can use $\LaTeX$.

The problem is as follows:

iBot...consider the ratio of the sum of the first n squares divided by the sum of the first n triangular numbers...now consider the limit of this ratio as n goes to infinity...what do you get?

Click to expand...

In order to solve this problem, we may rely on the following formulas:

$\displaystyle \sum_{k=1}^n\left(k\right)=\frac{n(n+1)}{2}\tag{1}$

$\displaystyle \sum_{k=1}^n\left(k^2\right)=\frac{n(n+1)(2n+1)}{6}\tag{2}$

These formulas can either be proved via induction or derived by solving inhomogeneous difference equations (and I will be glad to do either/both if requested).

Okay, so we already have the sum of the first $n$ squares, as given in (2)...and (1) gives us the triangular numbers, and so we need to compute:

$\displaystyle S=\sum_{k=1}^n\left(\frac{k(k+1)}{2}\right)$

Using the following properties of summations:

$\displaystyle \sum_{k=a}^b\left(c\cdot f(k)\right)=c\cdot\sum_{k=a}^b\left(f(k)\right)$

$\displaystyle \sum_{k=a}^b\left(f(k)\pm g(k)\right)= \sum_{k=a}^b\left(f(k)\right)\pm \sum_{k=a}^b\left(g(k)\right)$

We may now state:

$\displaystyle S=\frac{1}{2}\left(\sum_{k=1}^n\left(k^2\right)+ \sum_{k=1}^n\left(k\right)\right)$

Using (1) and (2), we obtain:

$\displaystyle S= \frac{1}{2}\left(\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}\right)= \frac{n(n+1)}{12}\left((2n+1)+3\right)= \frac{n(n+1)(n+2)}{6}$

And so, the ratio spoken of in the problem, which we'll call $R$, is:

$\displaystyle R=\frac{n(n+1)(2n+1)}{6}\cdot\frac{6}{n(n+1)(n+2)}=\frac{2n+1}{n+2}$

Now, to consider the ratio when $n$ grows without bound, we need to consider the following limit:

$\displaystyle L=\lim_{n\to\infty}\frac{2n+1}{n+2}= \lim_{n\to\infty}\frac{2+\dfrac{1}{n}}{1+\dfrac{2}{n}}= \frac{2+0}{1+0}=2$

And so the answer here is...$2$.