This question was uploaded on 02/12/21 on social media accounts.
Solution:
\[f(x) = \frac1{2^x} + \frac1{3^x} + ..... + \frac1{4000^x} = \sum_{n=2}^{4000}\frac1{n^x}\]
Let sum of f(1), f(2), f(3), ....., is 'S'.
\[S = f(2)+f(3)+f(4)+..... = \sum_{x=2}^{\infty}\left(\sum_{n=2}^{4000}\frac1{n^x}\right)\]
\[S = \sum_{x=2}^{\infty}\frac1{2^x} + \sum_{x=2}^{\infty}\frac1{3^x} + ....+\sum_{x=2}^{\infty}\frac1{4000^x}\]
Here, all terms are in infinite G.P.
\[\sum_{x=2}^{\infty}\frac1{a^x} = \frac{1/a^2}{1-1/a} = \frac1{a(a-1)} = \frac1{a-1} - \frac1{a}\]
\[\Rightarrow S = \left(\frac1{2-1} - \frac1{2}\right) + \left(\frac1{3-1} - \frac1{3}\right) + ....+\left(\frac1{4000-1} - \frac1{4000}\right)\]
\[\Rightarrow S = \left(\frac11 - \frac1{2}\right) + \left(\frac1{2} - \frac1{3}\right) + ....+\left(\frac1{3999} - \frac1{4000}\right)\]
It is a example of telescopic summation where only first and last term will remain and all other terms will cancel.
\[\Rightarrow S = 1 - \frac1{4000} = \frac{3999}{4000}\]
