limn→∞ 1+24+34+…+n4n5−limn→∞ 13+23+33+…+n3n5, is

A
$\frac{1}{5}$
B
$\frac{1}{30}$
C
$0$
D
$\frac{1}{4}$

Solution:

we have,

$\begin{array}{l} lim_{n \to \infty} \frac{1^{4} + 2^{4} + \dots + n^{4}}{n^{5}} - lim_{n \to \infty} \frac{1^{3} + 2^{3} + \dots + n^{3}}{n^{5}} \\ = lim_{n \to \infty} \frac{n (n + 1) (2 n + 1) (3 n^{2} + 3 n - 1)}{30 n^{5}} - lim_{n \to \infty} \frac{n^{2} (n + 1)^{2}}{4 n^{5}} \\ = lim_{n \to \infty} \frac{1}{30} (1 + \frac{1}{n}) (2 + \frac{1}{n}) (3 + \frac{3}{n} - \frac{1}{n^{2}}) - \frac{1}{4} lim_{n \to \infty} \frac{1}{n} {(1 + \frac{1}{n})}^{2} \\ = \frac{6}{30} - 0 = \frac{1}{5} \end{array}$

$lim_{n \to \infty} \frac{1 + 2^{4} + 3^{4} + \dots + n^{4}}{n^{5}} - lim_{n \to \infty} \frac{1^{3} + 2^{3} + 3^{3} + \dots + n^{3}}{n^{5}}$ , is

Solution:

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