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Consider, 

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

$\Rightarrow f\left(n\right)=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+.....+\frac{1}{n^2}$

The expression $\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+.....+\frac{1}{n^2}$  is the sum of the squares of the reciprocals of the positive integers from 1 to n. This is also known as the "sum of squares of reciprocals" or the "Basel problem". There is no closed-form formula to represent this sum, but it has been proven to converge to a specific value. This value is denoted by the Greek letter $\eta$ and is approximately equal to 1.64493. 

The exact value of $\eta$ is given by the following infinite series:

$\eta =\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+.....=\frac{{\mathrm{\pi }}^2}{6}$