Verified answer

[tex] \Large\textrm{\underline{\underline{Question}}}[/tex]

[tex] \large\rm \gray{ If \: f(x) = \frac{1}{ {2}^x } + \frac{1}{ {3}^{x} } + \frac{1}{ {4}^{x} } + \cdots + \frac{1}{ {4000}^x }} \\ \rm \gray{find \: f(2) + f(3) + f(4) + \cdots }[/tex]

[tex] \Large\textrm{\underline{\underline{Answer}}}[/tex]

[tex]\displaystyle f(2)+f(3)+f(4)+\cdots=\dfrac{3999}{4000}[/tex]

[tex] \Large\textrm{\underline{\underline{Explanation}}}[/tex]

[tex]\begin{gathered}\begin{aligned} &f(2)=\dfrac{1}{2^{2}}+\dfrac{1}{3^{2}}+\dfrac{1}{4^{2}}+\cdots+\dfrac{1}{4000^{2}} \\ &f(3)=\dfrac{1}{2^{3}}+\dfrac{1}{3^{3}}+\dfrac{1}{4^{3}}+\cdots+\dfrac{1}{4000^{3}} \\ &f(4)=\dfrac{1}{2^{4}}+\dfrac{1}{3^{4}}+\dfrac{1}{4^{4}}+\cdots+\dfrac{1}{4000^{4}} \\ &\cdots \end{aligned}\end{gathered} [/tex]

[tex] \therefore\displaystyle f(2)+f(3)+f(4)+\cdots=\sum^{\infty}_{n=2}\dfrac{1}{2^{n}}+\sum^{\infty}_{n=2}\dfrac{1}{3^{n}}+\sum^{\infty}_{n=2}\dfrac{1}{4^{n}}+\cdots+\sum^{\infty}_{n=2}\dfrac{1}{4000^{n}}[/tex]

[tex] \large\textrm{Infinite Geometric Series}[/tex]

[tex] \boxed{a_{n}=ar^{n-1},\ |r| < 1\Longrightarrow \displaystyle\sum^{\infty}_{n=1}a_{n}=\dfrac{a}{1-r}} [/tex]

[tex] \bf{Now,}[/tex]

[tex] \displaystyle \sum^{\infty}_{n=2}\dfrac{1}{2^{n}}+\sum^{\infty}_{n=2}\dfrac{1}{3^{n}}+\sum^{\infty}_{n=2}\dfrac{1}{4^{n}}+\cdots+\sum^{\infty}_{n=2}\dfrac{1}{4000^{n}} [/tex]

[tex] = \dfrac{\dfrac{1}{2^{2}}}{1-\dfrac{1}{2}}+\dfrac{\dfrac{1}{3^{2}}}{1-\dfrac{1}{3}}+\dfrac{\dfrac{1}{4^{2}}}{1-\dfrac{1}{4}}+\cdots+\dfrac{\dfrac{1}{4000^{2}}}{1-\dfrac{1}{4000}}[/tex]

[tex]=\dfrac{1}{2^{2}-2}+\dfrac{1}{3^{2}-3}+\dfrac{1}{4^{2}-4}+\cdots+\dfrac{1}{4000^{2}-4000}[/tex]

[tex]=\displaystyle\sum^{4000}_{n=2}\dfrac{1}{n(n-1)}[/tex]

[tex]=\displaystyle\sum^{4000}_{n=2}\dfrac{1}{n(n-1)}[/tex]

[tex]=1-\dfrac{1}{4000}[/tex]

[tex]=\dfrac{3999}{4000}[/tex]

[tex] \bf[And \: this \: is \: the \: required \: answer!][/tex]