1. ## Sigma Notation

I'm sorry if you have a diffuclty reading this. I am not sure how i would go about displaying this in HTML

well

above the Sigma there is an (n) and below (i=1) on the right side of the sigma this is displayed (1+ 2i/n) (2/n). Let s(n) = the above problem

Find the limit of s(n) as n -> infinity

How do I solve this problem? Could someone help me out step by step please?

I'm sorry if you have a diffuclty reading this. I am not sure how i would go about displaying this in HTML

well

above the Sigma there is an (n) and below (i=1) on the right side of the sigma this is displayed (1+ 2i/n) (2/n). Let s(n) = the above problem

Find the limit of s(n) as n -> infinity

How do I solve this problem? Could someone help me out step by step please?
$\sum_{i=1}^n(1+\frac{2i}{n})(\frac{2}{n})=\frac{2} {n}\left[\sum_{i=1}^n1+\frac{2}{n}\sum_{i=1}^ni\right]$

Can you take it from here?

3. Originally Posted by VonNemo19
$\sum_{i=1}^n(1+\frac{2i}{n})(\frac{2}{n})=\frac{2} {n}\left[\sum_{i=1}^n1+\frac{2}{n}\sum_{i=1}^ni\right]$

Can you take it from here?
not to be a pain, but not quite

not to be a pain, but not quite
Well, do you see what I have done thus far?

5. VonNemo19's expression is the result of factoring; the $n$ is an upper limit, not an index, so it can come out. Now you should know both of these identities:

$\sum_{i=1}^n 1 = n$ (trivial), and

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

The second identity can be proved by induction, or, better yet, by the ingenius method of Gauss, letting $s$ be the sum:

s = 1 + 2 + 3 + ... + n
s = n + ... + 3 + 2 + 1

Each line has n terms, and the terms of each line, added together, equal n+1. Hence 2s = n(n+1), or s= n(n+1)/2.

Using these identities, you should be able to solve the problem easily, letting n approach infinity.

6. I did the problem out and came out with the number 2, is this correct?

I did the problem out and came out with the number 2, is this correct?
4 is correct.

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

You can see that $\frac{2}{n}$ will be a factor of every term, so therefore we can pull it out in front of the sigma symbol

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

Now, understand that if I were to sum the remaining terms together, they would expand together. Heres what I mean...

$\sum_{i=1}^n(1+\frac{2i}{n})=(1+\frac{2(1)}{n})+(1 +\frac{2(2)}{n})+(1+\frac{2(3)}{n})+...+(1+\frac{2 (n)}{n})$

So, you can see how I could rearange the terms to be like this...

$\sum_{i=1}^n(1+\frac{2i}{n})=[1+1+1+...+1]+\left[\frac{2(1)}{n}+\frac{2(2)}{n}+\frac{2(3)}{n}+...+\ frac{2(n)}{n}\right]$

So, it's like I have two sums

One is $\sum_{i=1}^n1$, and the other is $\sum_{i=1}^n\frac{2i}{n}$.

So, then this gives $\frac{2}{n}\sum_{i=1}^n(1+\frac{2i}{n})=\frac{2}{n }\left[\sum_{i=1}^n1+\sum_{i=1}^n\frac{2i}{n}\right]$

Now notice the second term inside the brackets. I can factor $\frac{2}{n}$ out

$\frac{2}{n}\sum_{i=1}^n(1+\frac{2i}{n})=\frac{2}{n }\left[\sum_{i=1}^n1+\frac{2}{n}\sum_{i=1}^ni\right]$

Now from the formulas

$\sum_{i=1}^nk=nk$. and $\sum_{i=1}^ni=\frac{n(n+1)}{2}$ we get

$\frac{2}{n}\left[\sum_{i=1}^n1+\frac{2}{n}\sum_{i=1}^ni\right]=\frac{2}{n}\left[n+\frac{2}{n}\cdot\frac{n(n+1)}{2}\right]=\frac{2}{n}[n+(n+1)]=\frac{2}{n}(2n+1)=4+\frac{2}{n}$

Taking the limit

$\lim_{n\to\infty}\sum_{i=1}^n(1+\frac{2i}{n})(\fra c{2}{n})=\lim_{n\to\infty}(4+\frac{2}{n})=4$

8. omg I don't get it . Help please.

omg I don't get it . Help please.
I updated my last post. I hope you find it helpful.

omg I don't get it . Help please.
Please specify where, in the complete worked solution, you are getting lost.

Thank you!

11. Here's a somewhat more compact working...

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

$=\frac{2}{n}\left[n+\left(\frac{2}{n}\cdot \frac{1}{2}n(n+1)\right)\right]$ (by the identities given in my previous post)

$=\frac{2}{n}\left(2n+1\right)$ (algebraic simplification)

$=4 + \frac{2}{n} \rightarrow 4 + 0 = 4 \text { as } n \rightarrow \infty$ (further simplification and taking the limit.)