Question

$\underset{k\to \infty }{\lim }\left(\frac{1^3+2^3+\dots \dots .+k^3}{k^4}\right)=$

• A. $0$
• B. $2$
• C. $\frac{1}{3}$
• D. $\infty$
• E. $\frac{1}{4}$

Answer 1

The correct option is E

$\frac{1}{4}$

Explanation for the correct option:

Finding the value of the function on applying the limits:

Given,

$\begin{array}{rcl}\underset{k\to \infty }{\lim }\left(\frac{1^3+2^3+\dots \dots .+k^3}{k^4}\right)& =& \underset{k\to \infty }{\lim }\frac{{\left[{\displaystyle \frac{k(k+1)}{2}}\right]}^2}{k^4}\left[\because 1^3+2^3+\dots \dots .+n^3={\left(\frac{n\left(n+1\right)}{2}\right)}^2\right]\\ & =& \underset{k\to \infty }{\lim }\frac{k^2{(k+1)}^2}{4k^4}\\ & =& \frac{1}{4}\underset{k\to \infty }{\lim }\frac{{(k+1)}^2}{k^2}\\ & =& \frac{1}{4}\underset{k\to \infty }{\lim }{\left(\frac{k+1}{k}\right)}^2\end{array}$

As$k\to \infty ,\frac{1}{k}\to 0$

$\underset{k\to \infty }{\lim }\left(\frac{1^3+2^3+\dots \dots .+k^3}{k^4}\right)=\frac{1}{4}\underset{k\to \infty }{\lim }{\left(1+\frac{1}{k}\right)}^2$

Applying limits,

$\begin{array}{rcl}\underset{k\to \infty }{\lim }\left(\frac{1^3+2^3+\dots \dots .+k^3}{k^4}\right)& =& \frac{1}{4}{(1+0)}^2\\ & =& \frac{1}{4}{\left(1\right)}^2\\ & =& \frac{1}{4}\end{array}$

Therefore, the correct answer is option (E).