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Q. 52

Expert-verifiedFound in: Page 326

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

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

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

Given :

$\underset{n\to \infty}{\mathrm{lim}}\sum _{k=1}^{n}\frac{{k}^{3}}{{n}^{4}+n+1}$$\underset{n\to \infty}{\mathrm{lim}}\sum _{k=1}^{n}\frac{{k}^{3}}{{n}^{4}+n+1}\phantom{\rule{0ex}{0ex}}=\underset{n\to \infty}{\mathrm{lim}}\frac{1}{{n}^{4}+n+1}\sum _{k=1}^{n}{k}^{3}\phantom{\rule{0ex}{0ex}}=\underset{n\to \infty}{\mathrm{lim}}\frac{1}{{n}^{4}+n+1}{\left(\frac{n(n+1)}{2}\right)}^{2}\phantom{\rule{0ex}{0ex}}=\underset{n\to \infty}{\mathrm{lim}}\frac{1}{{n}^{4}+n+1}\left(\frac{{n}^{2}{(n+1)}^{2}}{4}\right)\phantom{\rule{0ex}{0ex}}=\underset{n\to \infty}{\mathrm{lim}}\frac{1}{{n}^{4}+n+1}\left(\frac{{n}^{2}({n}^{2}+2n+1)}{4}\right)\phantom{\rule{0ex}{0ex}}=\underset{n\to \infty}{\mathrm{lim}}\frac{{n}^{4}\left(1+\frac{2}{n}+\frac{1}{{n}^{2}}\right)}{4{n}^{4}\left(1+\frac{1}{{n}^{3}}+\frac{1}{{n}^{4}}\right)}\phantom{\rule{0ex}{0ex}}=\underset{n\to \infty}{\mathrm{lim}}\frac{\left(1+\frac{2}{n}+\frac{1}{{n}^{2}}\right)}{4\left(1+\frac{1}{{n}^{3}}+\frac{1}{{n}^{4}}\right)}\phantom{\rule{0ex}{0ex}}=\frac{1+0+0}{4(1+0+0)}\phantom{\rule{0ex}{0ex}}=\frac{1}{4}\phantom{\rule{0ex}{0ex}}$

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