Q. 48 - 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 

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

Step-by-Step Solution

Answer

The limit of the sum is infinite.

Step 1. Given information

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

Step 2. Find the limit of the sum.

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