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Answers without the blur. Sign up and see all textbooks for free! Q. 48

Expert-verified Found in: Page 326 ### Calculus

Book edition 1st
Author(s) Peter Kohn, Laura Taalman
Pages 1155 pages
ISBN 9781429241861 # 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 $\underset{n\to \infty }{\mathrm{lim}}\sum _{k=1}^{n}\left({k}^{2}+k+1\right)$

The limit of the sum is infinite.

See the step by step solution

## Step 1. Given information

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

## Step 2. Find the limit of the sum.

$\underset{n\to \infty }{\mathrm{lim}}\sum _{k=1}^{n}{k}^{2}+k+1\phantom{\rule{0ex}{0ex}}=\underset{n\to \infty }{\mathrm{lim}}\left(\underset{k=1}{\overset{n}{\sum {k}^{2}}}+\sum _{k=1}^{n}k+\sum _{k=1}^{n}1\right)\phantom{\rule{0ex}{0ex}}=\underset{n\to \infty }{\mathrm{lim}}\left(\frac{n\left(n+1\right)\left(2n+1\right)}{6}+\frac{n\left(n+1\right)}{2}+n\right)\phantom{\rule{0ex}{0ex}}=\underset{n\to \infty }{\mathrm{lim}}\left(\frac{n\left(n+1\right)\left(2n+1\right)+3n\left(n+1\right)+6n}{6}\right)\phantom{\rule{0ex}{0ex}}=\underset{n\to \infty }{\mathrm{lim}}\left(\frac{n\left(2{n}^{2}+3n+1\right)+3{n}^{2}+3n+6n}{6}\right)\phantom{\rule{0ex}{0ex}}=\underset{n\to \infty }{\mathrm{lim}}\left(\frac{2{n}^{3}+3{n}^{2}+n+3{n}^{2}+3n+6n}{6}\right)\phantom{\rule{0ex}{0ex}}=\underset{n\to \infty }{\mathrm{lim}}\left(\frac{2{n}^{3}+6{n}^{2}+10n}{6}\right)\phantom{\rule{0ex}{0ex}}=\infty$ ### Want to see more solutions like these? 