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 \[\left( k + 1 \right)^5 - k^5 = 5 k^4 + 10 k^3 + 10 k^2 + 5k + 1\]

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

\[\left( n + 1 \right)^5 - 1 = 5 \sum^n_{k = 1} k^4 + 10 \sum^n_{k = 1} k^3 + 10 \sum^n_{k = 1} k^2 + 5 \sum^n_{k = 1} k + \sum^n_{k = 1} 1\]
\[ \Rightarrow n^5 + 5 n^4 + 10 n^3 + 10 n^2 + 5n = 5 \sum^n_{k = 1} k^4 + \frac{10 n^2 \left( n + 1 \right)^2}{4} + \frac{10n\left( n + 1 \right)\left( 2n + 1 \right)}{6} + \frac{5n\left( n + 1 \right)}{2} + n\]
\[ \Rightarrow 5 \sum^n_{k = 1} k^4 = n^5 + 5 n^4 + 10 n^3 + 10 n^2 + 4n - \frac{5 n^2 \left( n + 1 \right)^2}{2} - \frac{5n\left( n + 1 \right)\left( 2n + 1 \right)}{3} - \frac{5n\left( n + 1 \right)}{2}\]
\[ \Rightarrow 5 \sum^n_{k = 1} k^4 = n^5 + \frac{5 n^4}{2} + \frac{5 n^3}{3} - \frac{n}{6}\]

This expression on further simplification gives \[\sum^n_{k = 1} k^4 = \frac{n\left( n + 1 \right)\left( 2n + 1 \right)\left( 3 n^2 + 3n - 1 \right)}{30}\] 

\[\therefore \lim_{n \to \infty} \frac{1^4 + 2^4 + 3^4 + . . . + n^4}{n^5} - \lim_{n \to \infty} \frac{1^3 + 2^3 + . . . + n^3}{n^5}\]
\[ = \lim_{n \to \infty} \frac{n\left( n + 1 \right)\left( 2n + 1 \right)\left( 3 n^2 + 3n - 1 \right)}{30 n^5} - \lim_{n \to \infty} \frac{n^2 \left( n + 1 \right)^2}{4 n^5}\]
\[ = \frac{1}{30} \lim_{n \to \infty} \left( 1 + \frac{1}{n} \right)\left( 2 + \frac{1}{n} \right)\left( 3 + \frac{3}{n} - \frac{1}{n^2} \right) - \frac{1}{4} \lim_{n \to \infty} \frac{1}{n} \left( 1 + \frac{1}{n} \right)^2 \]
\[ = \frac{1}{30} \times \left( 1 + 0 \right) \times \left( 2 + 0 \right) \times \left( 3 + 0 - 0 \right) - \frac{1}{4} \times 0 \left( \lim_{n \to \infty} \frac{1}{n} = \lim_{n \to \infty} \frac{1}{n^2} = . . . = 0 \right)\] 

\[= \frac{1}{30} \times 6 - 0\]
\[ = \frac{1}{5}\]