$$$\sum_{n=5}^{\infty} \frac{1}{n^{2}}$$$

求$$$\sum_{n=5}^{\infty} \frac{1}{n^{2}}$$$。

Split the series:

$${\color{red}{\left(\sum_{n=5}^{\infty} \frac{1}{n^{2}}\right)}}={\color{red}{\left(\sum_{n=1}^{\infty} \frac{1}{n^{2}} + \sum_{n=1}^{4} - \frac{1}{n^{2}}\right)}}$$

Since the bounds are finite, the number of terms is finite as well, and we just calculate the sum by summing up the terms.

$${\color{red}{\left(\sum_{n=1}^{4} - \frac{1}{n^{2}}\right)}}={\color{red}{\left(-1 - \frac{1}{4} - \frac{1}{9} - \frac{1}{16}\right)}}$$

$${\color{red}{\left(\sum_{n=1}^{4} - \frac{1}{n^{2}}\right)}} + \sum_{n=1}^{\infty} \frac{1}{n^{2}}={\color{red}{\left(- \frac{205}{144}\right)}} + \sum_{n=1}^{\infty} \frac{1}{n^{2}}$$

$$$\sum_{n=1}^{\infty} \frac{1}{n^{2}}$$$ is a known series.

It is $$$\sum_{n=1}^{\infty} n^{- n_{0}}=\zeta\left(n_{0}\right)$$$, $$$n_{0} > 1$$$ with $$$n_{0}=2$$$.

Therefore,

$$- \frac{205}{144} + {\color{red}{\left(\sum_{n=1}^{\infty} \frac{1}{n^{2}}\right)}}=- \frac{205}{144} + {\color{red}{\left(\frac{\pi^{2}}{6}\right)}}$$

Hence,

$$\sum_{n=5}^{\infty} \frac{1}{n^{2}}=- \frac{205}{144} + \frac{\pi^{2}}{6}$$

$$$\sum_{n=5}^{\infty} \frac{1}{n^{2}} = - \frac{205}{144} + \frac{\pi^{2}}{6}\approx 0.221322955737115$$$A