Basic question

[sickplaya]

I just started calculus class and i encounter this question

suppose s(n) = i=1 Σ n (1 + 2i/n)(2/n)
find the limit of s(n) as n approaches to infinity

Note: the i=1 is on the bottom of the sigma symbol and the "n" is on the top of the sigma symbol; sorry for not knowing how to type it properly

i tried this and got an answer of zero; and i am wondering is my answer correct

 

[galactus]

Is this your limit:

\(\displaystyle \L\\\lim_{n\to\infty}\sum_{i=1}^{n}\left(1+\frac{2i}{n}\right)\frac{2}{n}\)

If so, the answer is not 0.

 

[soroban]

Hello, sickplaya!

You cannot add a sequence of positive numbers and get zero . . .

Given: \(\displaystyle \L\:s(n) \;= \;\sum^n_{i=1}\left(1\,+\,\frac{2i}{n}\right)\cdot\frac{2}{n}\)

Find: \(\displaystyle \L\:\lim_{n\to\infty}s(n)\)

Do you know these formulas?

. . \(\displaystyle \L\sum^n_{i=1}i \;=\;\underbrace{1\,+\,1\,+\,1\,+\,\cdots\,+\,1}_{n\text{ terms}} \;=\;n\)

. . \(\displaystyle \L\sum^n_{i=1}i \;=\;1\,+\,2\,+\,3\,+\,\cdots\,+\,n\;=\;\frac{n(n\,+\,1)}{2}\)


We have: \(\displaystyle \L\:s(n)\;=\;\frac{2}{n}\sum^n_{i=1}\left(1 + \frac{2i}{n}\right) \;=\;\frac{2}{n}\left[\sum^n_{i=1}1\,+\,\sum^n_{i=1}\frac{2}{n}i\right]\)

. . \(\displaystyle \L=\;\frac{2}{n}\left[\sum^n_{i=1}1 \,+\,\frac{2}{n}\sum^n_{i=1}i\right] \;=\;\frac{2}{n}\left[n\,+\,\frac{2}{n}\cdot\frac{n(n\,+\,1)}{2}\right] \;=\;\frac{2}{n}\left(2n + 1\right)\)

. . \(\displaystyle \L=\;2\left(\frac{2n}{n}\,+\,\frac{1}{n}\right) \;=\;2\left(2 + \frac{1}{n}\right)\)


Take the limit: \(\displaystyle \L\:\lim_{n\to\infty}2\left(2 + \frac{1}{n}\right) \;=\;2(2\,+\,0) \;=\;4\)

 

[sickplaya]

Thanks

Thanks a lot buddy