Q. 49 - 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 

$\underset{n\to \infty }{\lim }\sum _{k=1}^n\frac{{\left(k+1\right)}^2}{n^3-1}$

Step-by-Step Solution

Answer

The limit of the sum is finite and it is $\frac{1}{3}$.

Step 1. Given information

$\underset{n\to \infty }{\lim }\sum _{k=1}^n\frac{{\left(k+1\right)}^2}{n^3-1}$

Step 2. Find the limit of the sum.

$$\begin{gathered} \underset{n\to \infty }{\lim }\sum _{k=1}^n\frac{{\left(k+1\right)}^2}{n^3-1} \\ =\underset{n\to \infty }{\lim }\frac{1}{n^3-1}\sum _{k=1}^n{(k+1)}^2 =\underset{n\to \infty }{\lim }\frac{1}{n^3-1}\sum _{k=1}^n\left(k^2+2k+1\right) \\ =\underset{n\to \infty }{\lim }\frac{1}{n^3-1}\left(\sum _{k=1}^n{k}^2+\sum _{k=1}^{n}2k+\sum _{k=1}^{n}1\right) \\ =\underset{n\to \infty }{\lim }\frac{1}{n^3-1}\left(\sum _{k=1}^n{k}^2+2\sum _{k=1}^{n}k+\sum _{k=1}^{n}1\right) \\ =\underset{n\to \infty }{\lim }\frac{1}{n^3-1}\left(\frac{n(n+1)(2n+1)}{6}+2\frac{n(n+1)}{2}+n\right) \\ =\underset{n\to \infty }{\lim }\frac{1}{n^3-1}\left(\frac{2n^3+3n^2+n}{6}+n^2+n+n\right) \\ =\underset{n\to \infty }{\lim }\frac{1}{n^3-1}\left(\frac{2n^3+3n^2+n+6n^2+12n}{6}\right) \\ =\underset{n\to \infty }{\lim }\left(\frac{2n^3+9n^2+13n}{6n^3-6}\right) \\ =\underset{n\to \infty }{\lim }\frac{n^3\left(2+\frac{9}{n}+\frac{13}{n^2}\right)}{n^3\left(6-{\displaystyle \frac{6}{n^3}}\right)} \\ =\underset{n\to \infty }{\lim }\frac{\left(2+\frac{9}{n}+\frac{13}{n^2}\right)}{\left(6-{\displaystyle \frac{6}{n^3}}\right)} \\ =\frac{2}{6} \\ =\frac{1}{3} \end{gathered}$$