Six Hundred Followers Problem

Calculus Level 2

$\lim_{n \to \infty} \left ( \frac {3}{1^2} + \frac {5}{1^2+2^2} + \frac {7}{1^2+2^2+3^2} + \ldots + \frac {2n+1}{1^2+ 2^2 + \ldots + n^2} \right ) = \ ?$

The answer is 6.

2 solutions

Pranjal Jain
Feb 21, 2015

$T_n=\dfrac{2n+1}{1^2+2^2+\cdots+n^2}\\=\dfrac{6(2n+1)}{n(n+1)(2n+1)}\\=\dfrac{6}{n(n+1)}\\=\dfrac{6}{n}-\dfrac{6}{n+1}$

$S=6-\dfrac{6}{2}+\dfrac{6}{2}-\dfrac{6}{3}+\cdots$

This is clearly a telescoping series.

$S=6$

You can also express the whole problem as a nested sum (albeit this may seem as abuse of notation to OP) as follows:

$\sum_{i=1}^\infty \left( \frac{2i+1}{\displaystyle \sum_{j=1}^i \left(j^2\right)} \right)=6\left\{\sum_{i=1}^\infty \left( \frac{1}{i}-\frac{1}{i+1}\right)\right\}$

Prasun Biswas - 6 years, 3 months ago

Priyesh Pandey
Feb 24, 2015

now its good this question should be of level 3

Six Hundred Followers Problem

The answer is 6.

2 solutions

0 pending reports