# Help with summations?

1. May 10, 2009

### klite

1. The problem statement, all variables and given/known data
Lim
n$$\rightarrow$$$$\infty$$ $$\sum(1+2i/n)^2(2/n)$$

2. Relevant equations
$$\sumi = n(n+1)/2$$
$$\sumi^2 = n(n+1)(2n+1)/6$$
$$\sum[n(n+1)/2]^2$$

3. The attempt at a solution
n$$\rightarrow$$$$\infty$$ $$\sum(1+2i/n)^2(2/n)$$

= n$$\rightarrow$$$$\infty$$ 2/n[$$\sum(1+4i/n+4i^2/n^2)]$$

= n$$\rightarrow$$$$\infty$$ 2/n (n) (n^2) $$\sum1$$ + $$4\sumi$$ + $$4\sumi^2$$

= n$$\rightarrow$$$$\infty$$ 2n^2 [n + 4[n(n+1)/2] + 2[n(n+1)(2n+1)/3]]

=...?

2. May 10, 2009

### Cyosis

Could you rewrite your post so things are readable? First of all what kind of sum is it. Does it run from n=0 to infinity? Is the 'i' in the sum the imaginary unit? In your relevant equations sections you've written down empty spaces being equal to two expressions that look like the values for the the sum of the first n natural numbers and the sum of the first n natural numbers squared. And in the attempt at a solution you're taking n in front of the sum, are you not summing over n, summing over i? If you're summing over n you can't take n in front of the sum.

Tip: Put the entire equations between $$brackets, not just the sum sign. You can specify boundaries like this, \sum_{n=0}^N, this sum runs from n=0 to n=N. 3. May 10, 2009 ### klite It runs from n=0 to infinity. The i is not the imaginary unit. Let me try to word the question this time: What is the sum of (1+2i/n)^2 (2/n) as n approaches infinity? Above the summation symbol is n and below the symbol is i=1. 4. May 10, 2009 ### Cyosis This suggests: [tex]\sum_{n=0}^\infty (1+\frac{2i}{n})^2 \frac{2}{n}$$
This suggests:
$$\sum_{i=1}^n (1+\frac{2i}{n})^2 \frac{2}{n}$$

Which one is it?

5. May 10, 2009

### klite

It's this one:

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

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

is limit as n $$\rightarrow\infty$$.

6. May 10, 2009

### Cyosis

Ah it starts to make sense now.

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

You can take 2/n in front in this case and write out the brackets, then break up the sum into each of its components.
hint: $$\sum_{i=1}^n (a_i+b_i)=\sum_{i=1}^n a_i+\sum_{i=1}^n b_i$$

Last edited: May 10, 2009
7. May 10, 2009

### klite

Ohh...I have no idea how to use the symbols, I'm new here :tongue:

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

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

= $$\lim_{n \to \infty} 2/n [\sum_{i=1}^n (1) + 4\sum_{i=1}^n (i) + 4\sum_{i^2}$$

= $$\lim_{n \to \infty} 2n^2 [n + 4(n(n+1)/2) + 4(n(n+1)(2n+1)/6)]$$

= $$\lim_{n \to \infty} 2n^2 [n + 2(n(n+1)) + 2(n(n+1)(2n+1)/3)]$$

This is as far as I got...I know the answer should be a fraction, but I don't realize how to move on to the next step from here.

8. May 10, 2009

### Cyosis

Something is going wrong with your notation. I'll do the first four steps and you check if you have the same on paper.

\begin{align*} \lim_{n \to \infty} \sum_{i=1}^n (1+\frac{2i}{n})^2 \frac{2}{n} & = \lim_{n \to \infty} \frac{2}{n} \sum_{i=1}^n (1+\frac{4 i^2}{n^2}+\frac{4i}{n}) \\ & =\lim_{n \to \infty} \frac{2}{n} \left( \sum_{i=1}^n 1+\sum_{i=1}^n \frac{4 i^2}{n^2}+\sum_{i=1}^n \frac{4i}{n} \right) \\ & = \lim_{n \to \infty} \frac{2}{n} \left( \sum_{i=1}^n 1+\frac{4}{n^2} \sum_{i=1}^n i^2+\frac{4}{n} \sum_{i=1}^n i \right) \end{align*}

Last edited: May 10, 2009
9. May 10, 2009

### klite

I have something similar to this...

$$\lim_{n \to \infty} n^2 [\sum_{i=1}^n (1) + 4\sum_{i=1}^n (i) + 4\sum_{i^2}i^2]$$

But your steps make it much more clearer, thank you.

Afterwards I just use these equations for 1, i, and i^2,
$$\sum_{i=1}^n 1 = 1*n = n$$

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

$$\sum_{i=1}^n i^2 = n^2(n+1)^2/4$$

correct?

10. May 10, 2009

### Cyosis

How did you get this?

The first sum is correct, the second sum is wrong there should be no division by n instead it should be division by 2. The third sum is also wrong. Check your relevant equations section, you have the correct expressions listed there.

11. May 10, 2009

### klite

I took 2/n out in front, after which I took the denominators n and n^2 out in front, and multiplied them by 2/n, yielding 2n^2 out in front. Then I took both 4s out from the sums, leaving

$$\sum_{i=1}^n 1$$

$$4\sum_{i=1}^n i$$

and $$4\sum_{i=1}^n i^2$$ inside the brackets, and 2n^2 outside the brackets.

Sorry, I meant divide it by 2 not n. And I don't understand how the third sum is wrong?

12. May 10, 2009

### Cyosis

Well the third sum is wrong, because of the following:

$$\sum_{i=1}^n i=1+2+3+4+...+n=n(n+1)/2 \Rightarrow n^2(n+1)^2/4 =(n(n+1)/2)^2=\left(\sum_{i=1}^n i \right)^2=(1+2+3+4+...+n)^2 \neq 1^2+2^2+3^2+4^2+...+n^2=\sum_{i=1}^n i^2$$

What you basically did is $(a+b)^2=a^2+b^2$, which is of course wrong. What confuses me though is that you've listed $\sum_{i=1}^n i^2=n(n+1)(1+2n)/6$ in your relevant equations section (which is correct). So I really don't see why this is suddenly posing a problem.

You can't take out 1/n^2 and 1/n, because they aren't present in all terms $ab+ac=a(b+c)$, but $ab+cd \neq ac(b+d)$.

Last edited: May 10, 2009
13. May 10, 2009

### klite

Oh, I see. Therefore the correct sum is (n(n+1)/2)^2?

We were writing them down in class, and what I got from my teacher was:

$$\sum_{i=1}^n i^2 = n(n+1)(2n+1)/6$$

$$\sum_{i=1}^n i^3 = (n(n+1)/2)^2 = n^2(n+1)^2/4$$

So where would I go from
$$\lim_{n \to \infty} \frac{2}{n} \left( \sum_{i=1}^n 1+\frac{4}{n^2} \sum_{i=1}^n i^2+\frac{4}{n} \sum_{i=1}^n i \right)$$?

14. May 10, 2009

### Cyosis

You should quote my edited line. There should be an i in each sum instead of an n, I have fixed that now.

No, look at my previous post again closely. It proves that $$\left(\sum_{i=1}^n i \right)^2 \neq \sum_{i=1}^n i^2$$. So you can't just take the value for $\sum_{i=1}^n i$ and then square it to obtain the expression for $\sum_{i=1}^n i^2$.

You know the value of all three sums in terms of n. Replace the sums with those expressions.

15. May 10, 2009

### klite

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

= $$\lim_{n \to \infty} \frac{2}{n} \left( n+\frac{2n(n+1)}{n^2} + \frac{2n(n+1)(2n+1)}{3n} \right)$$

= $$\lim_{n \to \infty} \frac{2}{n} \left( n+2+2n+\frac{2n(2n^2+3n+1)}{3n} \right)$$

= $$\lim_{n \to \infty} \frac{2}{n} \left( 3n+2+\frac{4n^3+6n^2+2n}{3n} \right)$$

= $$\lim_{n \to \infty} \frac{2}{n} \left( 3n+2+4n^3+2n+2n \right)$$

I know I'm doing something wrong...I can't find it? We've only just started learning summations and it's a concept I can't really grasp. Could you please show me the rest of the steps to find the answer, then I can do the rest of the summation homework with problems like these?

16. May 10, 2009

### Cyosis

The mistake you made is in going from step 1 to 2. You substitute $\sum_{i=1}^n i^2=n(n+1)/2$ and $\sum_{i=1}^n i=n(n+1)(1+2n) /6$. Compare this with your notes and you will see what is wrong.

[quote='klite]
I know I'm doing something wrong...I can't find it? We've only just started learning summations and it's a concept I can't really grasp. Could you please show me the rest of the steps to find the answer, then I can do the rest of the summation homework with problems like these?[/quote]

We will work our way to the correct answer step by step don't worry about it. It's important though that you do a part, if not most of the thinking so you truly understand what's going on. Therefore I am not going to give you the entire solution yet!

17. May 10, 2009

### klite

Alright, thank you!
I noticed the mistake - wrong sums:

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

= $$\lim_{n \to \infty} \frac{2}{n} \left( n + \frac{4}{n^2} \frac{2n(n+1)(2n+1)}{6} + \frac{4}{n} \frac{n(n+1)}{2} \right)$$

= $$\lim_{n \to \infty} \frac{2}{n} \left( n + \frac{2}{n^2} \frac{2n(n+1)(2n+1)}{3} + \frac{2}{n} n(n+1) \right)$$

= $$\lim_{n \to \infty} \frac{2}{n} \left( n + \frac{4n(n+1)(2n+1)}{3n^2} + \frac{2n(n+1)}{n} \right)$$

= $$\lim_{n \to \infty} \frac{2}{n} \left( n + \frac{4n(2n^2+3n+1}{3n^2} + {2n^2+2n}{n} \right)$$

= $$\lim_{n \to \infty} \frac{2}{n} \left( n + \frac{8n^3+12n^2+4n}{3n^2} + \frac{2n^2+2n}{n} \right)$$

= $$\lim_{n \to \infty} \frac{2}{n} \left( n + 8n^3 + 4 + 4n (+0)) \right)$$

I'm not sure whether the last step is correct.

Edit:

= $$\lim_{n \to \infty} \frac{2}{n} \left(\frac{8}{3} + 2 \right)$$?

Last edited: May 10, 2009
18. May 10, 2009

### Cyosis

What is that red 2 doing there? It shouldn't be there. Do you see why it shouldn't be there? Aside from that extra two in the second term all other steps up to the last one are correct. The last one however is not, how did you go from the second to last step to the last step?

Last edited: May 10, 2009
19. May 10, 2009

### klite

I was looking at an older incorrect step of the problem...it's another mistake. I meant:

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

$$\lim_{n \to \infty} \frac{2}{n} \left( n + \frac{4}{n^2} \frac{{\color{red}2}n(n+1)(2n+1)}{6} + \frac{4}{n} \frac{n(n+1)}{2} \right)$$

If I use n instead of 2n,

$$\lim_{n \to \infty} \frac{2}{n} \left( n + \frac{2n(2n^2+3n+1}{3n^2} + {2n^2+2n}{n} \right)$$

= $$\lim_{n \to \infty} \frac{2}{n} \left( n + \frac{2n(n+1)(2n+1)}{3n^2} + \frac{2n(n+1)}{n} \right)$$

= $$\lim_{n \to \infty} \frac{2}{n} \left( n + \frac{2n(2n^2+3n+1}{3n^2} + {2n^2+2n}{n} \right)$$

= $$\lim_{n \to \infty} \frac{2}{n} \left( n + \frac{4n^3+12n^2+4n}{3n^2} + \frac{2n^2+2n}{n} \right)$$

I multiply through the $$\frac{2}{n}$$ and get:

= $$\lim_{n \to \infty} \left(\frac{2n}{n} + \frac{8n^3......}{3n^3} + \frac{4n^2......}{n^2} \right)$$

= $$\frac{2}{1} + \frac{8}{3} + \frac{4}{1}$$

= $$\frac{8}{3} + 6$$ = $$\frac{26}{3}$$

20. May 10, 2009

Correct.