# How do you prove that sum of infinite series 1+1/4+1/9+............... is less than two?

Feb 23, 2017

See explanation...

#### Explanation:

${\sum}_{n = 1}^{\infty} \frac{1}{n} ^ 2 = 1 + {\sum}_{n = 2}^{\infty} \frac{1}{n} ^ 2 < 1 + {\sum}_{n = 2}^{\infty} \frac{1}{n \left(n - 1\right)}$

Then:

$\frac{1}{n \left(n - 1\right)} = \frac{n - \left(n - 1\right)}{n \left(n - 1\right)} = \frac{1}{n - 1} - \frac{1}{n}$

So:

${\sum}_{n = 2}^{N} \frac{1}{n \left(n - 1\right)} = {\sum}_{n = 2}^{N} \left(\frac{1}{n - 1} - \frac{1}{n}\right)$

$\textcolor{w h i t e}{{\sum}_{n = 2}^{N} \frac{1}{n \left(n - 1\right)}} = {\sum}_{n = 2}^{N} \frac{1}{n - 1} - {\sum}_{n = 2}^{N} \frac{1}{n}$

$\textcolor{w h i t e}{{\sum}_{n = 2}^{N} \frac{1}{n \left(n - 1\right)}} = 1 + {\sum}_{n = 3}^{N} \frac{1}{n - 1} - {\sum}_{n = 2}^{N - 1} \frac{1}{n} - \frac{1}{N}$

$\textcolor{w h i t e}{{\sum}_{n = 2}^{N} \frac{1}{n \left(n - 1\right)}} = 1 + \textcolor{red}{\cancel{\textcolor{b l a c k}{{\sum}_{n = 2}^{N - 1} \frac{1}{n}}}} - \textcolor{red}{\cancel{\textcolor{b l a c k}{{\sum}_{n = 2}^{N - 1} \frac{1}{n}}}} - \frac{1}{N}$

$\textcolor{w h i t e}{{\sum}_{n = 2}^{N} \frac{1}{n \left(n - 1\right)}} = 1 - \frac{1}{N}$

So:

${\sum}_{n = 2}^{\infty} \frac{1}{n \left(n - 1\right)} = {\lim}_{N \to \infty} \left(1 - \frac{1}{N}\right) = 1$

So:

${\sum}_{n = 1}^{\infty} \frac{1}{n} ^ 2 < 1 + {\sum}_{n = 2}^{\infty} \frac{1}{n \left(n - 1\right)} = 1 + 1 = 2$