Question

Evaluate: $\underset{n\to \infty }{\lim }\frac{1^4+2^4+3^4+...+n^4}{n^5}-\underset{n\to \infty }{\lim }\frac{1^3+2^3+...+n^3}{n^5}$

Answer 1

Consider the identity

${\left(k+1\right)}^5-k^5=5k^4+10k^3+10k^2+5k+1$ .....(1)

Putting k = 1, 2, 3,..., n in (1) and then adding the equations, we have

$$\begin{gathered} {\left(n+1\right)}^5-1=5\sum _{k=1}^n{k}^4+10\sum _{k=1}^n{k}^3+10\sum _{k=1}^n{k}^2+5\sum _{k=1}^{n}k+\sum _{k=1}^{n}1 \\ \Rightarrow n^5+5n^4+10n^3+10n^2+5n=5\sum _{k=1}^n{k}^4+\frac{10n^2{\left(n+1\right)}^2}{4}+\frac{10n\left(n+1\right)\left(2n+1\right)}{6}+\frac{5n\left(n+1\right)}{2}+n \\ \Rightarrow 5\sum _{k=1}^n{k}^4=n^5+5n^4+10n^3+10n^2+4n-\frac{5n^2{\left(n+1\right)}^2}{2}-\frac{5n\left(n+1\right)\left(2n+1\right)}{3}-\frac{5n\left(n+1\right)}{2} \\ \Rightarrow 5\sum _{k=1}^n{k}^4=n^5+\frac{5n^4}{2}+\frac{5n^3}{3}-\frac{n}{6} \end{gathered}$$

This expression on further simplification gives

$\sum _{k=1}^n{k}^4=\frac{n\left(n+1\right)\left(2n+1\right)\left(3n^2+3n-1\right)}{30}$


$$\begin{gathered} \therefore \underset{n\to \infty }{\lim }\frac{1^4+2^4+3^4+...+n^4}{n^5}-\underset{n\to \infty }{\lim }\frac{1^3+2^3+...+n^3}{n^5} \\ =\underset{n\to \infty }{\lim }\frac{n\left(n+1\right)\left(2n+1\right)\left(3n^2+3n-1\right)}{30n^5}-\underset{n\to \infty }{\lim }\frac{n^2{\left(n+1\right)}^2}{4n^5} \\ =\frac{1}{30}\underset{n\to \infty }{\lim }\left(1+\frac{1}{n}\right)\left(2+\frac{1}{n}\right)\left(3+\frac{3}{n}-\frac{1}{n^2}\right)-\frac{1}{4}\underset{\mathrm{n}\to \infty }{\lim }\frac{1}{n}{\left(1+\frac{1}{n}\right)}^2 \\ =\frac{1}{30}\times \left(1+0\right)\times \left(2+0\right)\times \left(3+0-0\right)-\frac{1}{4}\times 0\left(\underset{n\to \infty }{\lim }\frac{1}{n}=\underset{n\to \infty }{\lim }\frac{1}{n^2}=...=0\right) \end{gathered}$$
$$\begin{gathered} =\frac{1}{30}\times 6-0 \\ =\frac{1}{5} \end{gathered}$$