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

$\underset{n\to \infty }{\lim }\sum _{k=1}^n\frac{k^2+k+1}{n^2}$

Short answer

The limit of the sum is infinite.

Step-by-step solution

Step 1. Given information

$\underset{n\to \infty }{\lim }\sum _{k=1}^n\frac{k^2+k+1}{n^2}$

Step 2. Find the limit of the sum.

$$\begin{gathered} \underset{n\to \infty }{\lim }\sum _{k=1}^n\frac{k^2+k+1}{n^2} \\ =\underset{n\to \infty }{\lim }\sum _{k=1}^n\frac{k^2+k+1}{n^2} \\ =\underset{n\to \infty }{\lim }\frac{1}{n^2}\left(\underset{k=1}{\overset{n}{\sum k^2}}+\sum _{k=1}^{n}k+\sum _{k=1}^{n}1\right) \\ =\underset{n\to \infty }{\lim }\frac{1}{n^2}\left(\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}+n\right) \\ =\underset{n\to \infty }{\lim }\frac{n}{n^2}\left(\frac{(n+1)(2n+1)}{6}+\frac{(n+1)}{2}+1\right) \\ =\underset{n\to \infty }{\lim }\frac{1}{n}\left(\frac{(n+1)(2n+1)+3(n+1)+6}{6}\right) \\ =\underset{n\to \infty }{\lim }\frac{1}{n}\left(\frac{2n^2+n+2n+1+3n+3+6}{6}\right) \\ =\underset{n\to \infty }{\lim }\left(\frac{2n^2+6n+10}{6n}\right) \\ =\infty \end{gathered}$$