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

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

Short answer

The limit of the sum is finite and it is equal to$\frac{7}{3}$.

Step-by-step solution

Step 1. Given information

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

Step 2. Find limit of the sum.

$$\begin{gathered} \underset{n\to \infty }{\lim }\sum _{k=1}^n{\left(1+\frac{k}{n}\right)}^2.\frac{1}{n} \\ =\underset{n\to \infty }{\lim }\frac{1}{n}\sum _{k=1}^n{\left(1+\frac{k}{n}\right)}^2 \\ =\underset{n\to \infty }{\lim }\frac{1}{n}.\sum _{k=1}^n{\left(1\right)}^2+2.1.\frac{k}{n}+{\left(\frac{k}{n}\right)}^2 \\ =\underset{n\to \infty }{\lim }\frac{1}{n}.\sum _{k=1}^n\left(1+2\frac{k}{n}+\frac{k^2}{n^2}\right) \\ =\underset{n\to \infty }{\lim }\frac{1}{n}.\left(\sum _{k=1}^{n}1+\frac{2}{n}\sum _{k=1}^{n}k+\frac{1}{n^2}\sum _{k=1}^n{k}^2\right) \\ =\underset{n\to \infty }{\lim }\frac{1}{n}\left(n+\frac{2}{n}.\frac{n(n+1)}{2}+\frac{1}{n^2}.\frac{n(n+1)(2n+1)}{6}\right) \\ =\underset{n\to \infty }{\lim }\frac{1}{n}\left(n+n+1+\frac{(n+1)(2n+1)}{6n}\right) \\ =\underset{n\to \infty }{\lim }\left(\frac{6n(2n+1)+(n+1)(2n+1)}{6n^2}\right) \\ =\underset{n\to \infty }{\lim }\left(\frac{12n^2+6n+2n^2+3n+1}{6n^2}\right) \\ =\underset{n\to \infty }{\lim }\left(\frac{14n^2+9n+1}{6n^2}\right) \\ =\underset{n\to \infty }{\lim }\frac{n^2\left(14+{\displaystyle \frac{9}{n}}+{\displaystyle \frac{1}{n^2}}\right)}{6n^2} \\ =\underset{n\to \infty }{\lim }\frac{\left(14+{\displaystyle \frac{9}{n}}+{\displaystyle \frac{1}{n^2}}\right)}{6} \\ =\frac{14+0+0}{6} \\ =\frac{7}{3} \end{gathered}$$