Q. 52 - Calculus

Question

Determine which of the limit of sums in Exercises 47–52 are infinite and which are finite. For each limit of sums that is finite, compute its value   

$\underset{n\to \infty }{\lim }\sum _{k=1}^n\frac{k^3}{n^4+n+1}$

Step-by-Step Solution

Answer

The limit of the sum is finite and it is equal to $\frac{1}{4}$. 

Step 1. Given information

Given :

$\underset{n\to \infty }{\lim }\sum _{k=1}^n\frac{k^3}{n^4+n+1}$

Step 2. Find limit of the sum.

$$\begin{gathered} \underset{n\to \infty }{\lim }\sum _{k=1}^n\frac{k^3}{n^4+n+1} \\ =\underset{n\to \infty }{\lim }\frac{1}{n^4+n+1}\sum _{k=1}^n{k}^3 \\ =\underset{n\to \infty }{\lim }\frac{1}{n^4+n+1}{\left(\frac{n(n+1)}{2}\right)}^2 \\ =\underset{n\to \infty }{\lim }\frac{1}{n^4+n+1}\left(\frac{n^2{(n+1)}^2}{4}\right) \\ =\underset{n\to \infty }{\lim }\frac{1}{n^4+n+1}\left(\frac{n^2(n^2+2n+1)}{4}\right) \\ =\underset{n\to \infty }{\lim }\frac{n^4\left(1+{\displaystyle \frac{2}{n}}+{\displaystyle \frac{1}{n^2}}\right)}{4n^4\left(1+\frac{1}{n^3}+\frac{1}{n^4}\right)} \\ =\underset{n\to \infty }{\lim }\frac{\left(1+{\displaystyle \frac{2}{n}}+{\displaystyle \frac{1}{n^2}}\right)}{4\left(1+\frac{1}{n^3}+\frac{1}{n^4}\right)} \\ =\frac{1+0+0}{4(1+0+0)} \\ =\frac{1}{4} \end{gathered}$$