Question

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

Answer 1

Consider the identity

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

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

$(n+1{)}^5-1=5{\sum }_{k=1}^n{k}^4+10{\sum }_{k=1}^n{k}^3+10{\sum }_{k=1}^n{k}^2+5{\sum }_{k=1}^{n}k+{\sum }_{k=1}^{n}1$

$\Rightarrow n^5+5n^4+10n^3+10n^2+5n=5{\sum }_{k=1}^n{k}^4+\frac{10n^2(n+1{)}^2}{4}+\frac{10(n+1)(2n+1)}{6}+\frac{5n(n+1)}{2}+n$

$\Rightarrow 5{\sum }_{k=1}^n{k}^4=n^5+5n^4+10n^3+10n^2+4n-\frac{5n^2(n+1)}{2}-\frac{5n(n+1)(2n+1)}{3}-\frac{5n(n+1)}{2}$

$\Rightarrow 5{\sum }_{k=1}^n{k}^4=n^5+\frac{5n^4}{2}+\frac{5n^3}{3}-\frac{n}{6}$

This expression on further simplification gives

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

$\therefore {lim}_{n\to \infty }\frac{1^4+2^4+3^4+....+n^4}{n^5}-{lim}_{n\to \infty }\frac{1^3+2^3....+n^3}{n^5}$

$={lim}_{n\to \infty }\frac{n(n+1)(2n+1)(3n^2+3n-1)}{30}-{lim}_{n\to \infty }\frac{n^2(n+1{)}^2}{4n^5}$

$=\frac{1}{30}{lim}_{n\to \infty }(1+\frac{1}{n})(1+\frac{2}{n})(3+\frac{3}{n}-\frac{1}{n^2})-\frac{1}{4}{lim}_{n\to \infty }\frac{1}{n}{(1+\frac{1}{n})}^2$

$=\frac{1}{30}\times (1+0)\times (2+0)\times (3+0-0)-\frac{1}{4}\times 0({lim}_{n\to \infty }\frac{1}{n}={lim}_{n\to \infty }\frac{1}{n^2}=...=0)$

$=\frac{1}{30}\times 6-0$