Priscilla Johnston

2021-12-13

How do you find a formula for the sum n terms $\sum _{i=1}^{n}\left(1+\frac{i}{n}\right)\left(\frac{2}{n}\right)$ and then find the limit as $n\to \mathrm{\infty }$

Ben Owens

$\sum _{i=1}^{n}\left(1+\frac{i}{n}\right)\left(\frac{2}{n}\right)=\frac{3n+1}{n}$
$\underset{n\to \mathrm{\infty }}{lim}\sum _{i=1}^{n}\left(1+\frac{i}{n}\right)\left(\frac{2}{n}\right)=3$
Explanation
Let ${S}_{n}=\sum _{i=1}^{n}\left(1+\frac{i}{n}\right)\left(\frac{2}{n}\right)$
$\therefore {S}_{n}=\sum _{i=1}^{n}\left(\frac{2}{n}+\frac{2i}{{n}^{2}}\right)$
$\therefore {S}_{n}=\frac{2}{n}\sum _{i=1}^{n}\left(1\right)+\frac{2i}{{n}^{2}}\sum _{i=1}^{n}\left(i\right)$
And using the standard results: $\sum _{i=1}^{n}r=\frac{1}{2}n\left(n+1\right)$
We have:
${S}_{n}=\frac{2}{n}\left(n\right)=\frac{2}{{n}^{2}}\cdot \frac{1}{2}n\left(n+1\right)$
$\therefore {S}_{n}=2+\frac{n+1}{n}$
$\therefore {S}_{n}=\frac{\left(2n\right)+\left(n+1\right)}{n}$
$\therefore {S}_{n}=\frac{3n+1}{n}$
Now we examine the behaviour of ${S}_{n}$ as $n\to \mathrm{\infty }$
We have:
${S}_{n}=\frac{3n+1}{n}$
$\therefore {S}_{n}=3+\frac{1}{n}$
$\therefore \underset{n\to \mathrm{\infty }}{lim}{S}_{n}=\underset{n\to \mathrm{\infty }}{lim}\left\{3+\frac{1}{n}\right\}$
$\therefore \underset{n\to \mathrm{\infty }}{lim}{S}_{n}=\underset{n\to \mathrm{\infty }}{lim}\left(3\right)-\underset{n\to \mathrm{\infty }}{lim}\left(\frac{1}{n}\right)$
And as $\frac{1}{n}\to 0$ as $n\to \mathrm{\infty }$ we have;