1. ## Area?

Find a formula for for sum of n terms.Use the formula to find limit as n approaches infinity.

sigma notation start i=1 end n

((1+(i/n))((2/n))

Would anyone kindly help me with this problem.

2. ## Re: Area?

I guess you mean this:
$\sum_{i=1}^{n} \frac{1+\frac{i}{n}}{\frac{2}{n}}$
?
Write it as:
$\sum_{i=1}^{n} \frac{1+\frac{i}{n}}{\frac{2}{n}}=\sum_{i=1}^{n} \frac{\frac{n+i}{n}}{\frac{2}{n}}=\sum_{i=1}^{n} \frac{n+i}{2}$

Do you recognize a 'special' serie? ...

3. ## Re: Area?

Originally Posted by homeylova223
Find a formula for for sum of n terms.Use the formula to find limit as n approaches infinity.

sigma notation start i=1 end n

((1+(i/n))((2/n))

Would anyone kindly help me with this problem.

It is Riemann's upper sum of function $f(x)=1+x/2$ on $[0,2]$.

$\int_0^2 (1+x/2) \ dx =3$

4. ## Re: Area?

Originally Posted by homeylova223
Find a formula for for sum of n terms.Use the formula to find limit as n approaches infinity.

sigma notation start i=1 end n

((1+(i/n))((2/n))

Would anyone kindly help me with this problem.
This is a Riemann sum for:

$I=\int_0^2 (1+x/2)\; dx$

If you have arrived at the limit from the Riemann sum you should say so, otherwise you will probably get the integral as the answer for the limit rather than the sum of the finite series followed by limiting.

CB

5. ## Re: Area?

Originally Posted by Siron
I guess you mean this:
$\sum_{i=1}^{n} \frac{1+\frac{i}{n}}{\frac{2}{n}}$
?
That should be:

$\sum_{i=1}^{n} \left(1+\frac{i}{n}\right)\frac{2}{n}}=2+\frac{2}{ n^2}\sum_{i=1}^n} i$

CB

6. ## Re: Area?

Originally Posted by CaptainBlack
That should be:

$\sum_{i=1}^{n} \left(1+\frac{i}{n}\right)\frac{2}{n}}=2+\frac{2}{ n^2}\sum_{i=1}^n} i$

CB
Yes, offcourse, I red it wrong.

7. ## Re: Area?

I do not know how to explain I am trying to find the sum of the finite series. But to do that would I apply the interval Zartathrusta and you gave me?

8. ## Re: Area?

Look at the post 5 of Captain Black. Calculate $\lim_{n \to \infty}$ of the sum that will give you the area.
Afterwards compare with post 3 (the integral) of Also sprach zarathrusta, you'll come to the same result.