# Limit of a term involvin sum

• Aug 20th 2013, 03:13 PM
MathoMan
Limit of a term involvin sum
Ok guys, it seems that I have become rusty since I cant find an apropriate way to tackle the following limit:
$\displaystyle \lim \limits_{n\to \infty}\frac{1}{n^2}\sum\limits_{i=1}^{n}i\cdot e^{1+\left(\frac{i}{n}\right)^2}$. (Wondering)

I really would appreciate any advice.

So far (I think) I have found both upper and lower boundary on the sum (probably not the best one) but thats all I got at the moment. Here goes...
$\displaystyle e^1\leq e^{1+\left(\frac{i}{n}\right)^2}\leq e^2, \qquad \forall n\in \mathbf{N}, i\in \mathbf{N},i\leq n.$

$\displaystyle \sum\limits_{i=1}^{n}i\cdot e\leq \sum\limits_{i=1}^{n}i\cdot e^{1+\left(\frac{i}{n}\right)^2}\leq \sum\limits_{i=1}^{n}i\cdot e^2$

$\displaystyle e\cdot \sum\limits_{i=1}^{n}i \leq \sum\limits_{i=1}^{n}i\cdot e^{1+\left(\frac{i}{n}\right)^2}\leq e^2\cdot \sum\limits_{i=1}^{n}i$

$\displaystyle e\cdot \frac{n(n+1)}{2} \leq \sum\limits_{i=1}^{n}i\cdot e^{1+\left(\frac{i}{n}\right)^2}\leq e^2\cdot \frac{n(n+1)}{2}$

$\displaystyle \frac{e}{2}\cdot n^2 +\frac{e}{2}n\leq \sum\limits_{i=1}^{n}i\cdot e^{1+\left(\frac{i}{n}\right)^2}\leq \frac{e^2}{2}\cdot n^2 +\frac{e^2}{2}n$

Now it seems one can conlude that

$\displaystyle \lim\limits_{n\to\infty}\frac{1}{n^2}\cdot \left( \frac{e}{2}\cdot n^2 +\frac{e}{2}n\right) \leq \lim \limits_{n\to \infty}\frac{1}{n^2}\sum\limits_{i=1}^{n}i\cdot e^{1+\left(\frac{i}{n}\right)^2}\leq \lim\limits_{n\to \infty} \frac{1}{n^2}\cdot\left(\frac{e^2}{2}\cdot n^2 +\frac{e^2}{2}n\right)$.

That means that $\displaystyle \frac{e}{2}\leq \lim \limits_{n\to \infty}\frac{1}{n^2}\sum\limits_{i=1}^{n}i\cdot e^{1+\left(\frac{i}{n}\right)^2}\leq \frac{e^2}{2}$.

So even if I'm right with all the above (which is not necessarily the case (Happy) ) , that still doesn't give me the exact value of the limit, so I'm pretty much stuck. I hope it is something trivial that I'm overlooking and that someone can direct me, give me a push, towards the solution.

PS. I have tried using Wolfram Mathematica and Wolfram Alpha but they both went numb on this one. So I evaluated the expression under the limit operator for n=100000 and got the number 2.335424 and also for n=1000000 in which case the outcome was 2.33549083. So I'm pretty much out of ideas.(Doh)
• Aug 21st 2013, 01:04 PM
Lord Voldemort
Re: Limit of a term involvin sum
I've worked out e*(e-1) as an upper bound, which curiously appears to be twice the value of the sequence for large n. Thus I think the limit is e*(e-1)/2, but I'm having trouble proving it. Maybe this will help you.
• Aug 21st 2013, 08:52 PM
MathoMan
Re: Limit of a term involvin sum
I'm still with this one. Its 7am here now, but tonight I'll see into it. Thank you very much for your response Lord Voldemort (where are the others, I wonder), I really appreciate it.
• Aug 22nd 2013, 10:47 AM
Lord Voldemort
Re: Limit of a term involvin sum
I was hoping someone could enlighten us, but I'll show you how I arrived to a better upper bound of e:

$\displaystyle \lim \limits_{n\to \infty}\frac{1}{n^2}\sum\limits_{i=1}^{n}i\cdot e^{1+\left(\frac{i}{n}\right)^2}$

Let $\displaystyle a_n = \frac{1}{n^2}\sum\limits_{i=1}^{n}i\cdot e^{1+\left(\frac{i}{n}\right)^2} = \frac{e}{n^2}\sum\limits_{i=1}^{n}i\cdot e^{\left(\frac{i}{n}\right)^2}$ ; note that $\displaystyle i \le n$ so we have $\displaystyle \left(\frac{i}{n}\right) \ge \left(\frac{i}{n}\right)^2$ and thus $\displaystyle e^{\left(\frac{i}{n}\right)} \ge e^{\left(\frac{i}{n}\right)^2}$ so we can say:

$\displaystyle a_n \le \frac{e}{n^2} \sum\limits_{i=1}^{n}i\cdot {\left(e^{\frac{1}{n}}\right)^{i}$ which has a nice representation (see the second one: List of mathematical series - Wikipedia, the free encyclopedia ). Thus:

$\displaystyle \frac{e}{n^2} \sum\limits_{i=1}^{n}i\cdot {\left(e^{\frac{1}{n}}\right)^{i} = \frac{e}{n^2} \cdot e^{\frac{1}{n}} \cdot \frac{1-(n+1) \cdot e + n \cdot e^{\frac{n+1}{n}}}{(e^{\frac{1}{n}}-1)^2} \to e$

The approach I tried before that led to $\displaystyle e \cdot (e-1)$ as an upper bound was similar except I used the fact that $\displaystyle \frac{e}{n^2} \sum\limits_{i=1}^{n}i\cdot {\left(e^{\frac{1}{n}}\right)^{i} \le \frac{e}{n} \sum\limits_{i=0}^{n}{\left(e^{\frac{1}{n}}\right) ^{i}$
• Aug 22nd 2013, 11:36 AM
MathoMan
Re: Limit of a term involvin sum
You really went the extra mile with your reply and I thank you for it. And as you mentioned in your previous reply, everything points in the direction of e*(e-1)/2 being the correct answer with the only 'minor' problem of giving a correct mathematical proof.
• Aug 23rd 2013, 12:00 PM
Lord Voldemort
Re: Limit of a term involvin sum
If you figure out how to prove this, please follow up in a post. I'm curious now. Where did this problem come from?
• Aug 24th 2013, 02:45 AM
twizter
Re: Limit of a term involvin sum
Here's how I did it. First pull out the e from $\displaystyle e^{1 + (i/n)^2}$ outside of the summation. Now expand $\displaystyle e^{(i/n)^2}$ using the Taylor series expansion for e^x. Now multiply this by$\displaystyle i/n^2$. You get $\displaystyle \sum\limits_{i=1}^n (i/n^2 + i^3/n^4 + \frac{1}{2!} \frac{i^5}{n^6} + ...)$. Now consider $\displaystyle \sum\limits_{i=1}^n i^k$ . The leading term of this is$\displaystyle n^k^+^1/(k+1)$. Now this is the only term that matters in the limit since it is being divided by $\displaystyle n^{k+1}$. So it reduces to$\displaystyle e (1/2 + 1/4 + 1/(2!*6) + 1/(3!*8) + ..)$. Factoring out the 1/2 from this we get $\displaystyle (e/2)*(1 + 1/2 + 1/(2!*3) + 1/(3!*4) + ... )$ which is nothing but $\displaystyle (e/2)*(1 + 1/2! + 1/3! + 1/4! + ... )$. This sum is equal to e -1. So the answer is $\displaystyle e*(e-1)/2$. Hope this helps (Wink)
• Aug 24th 2013, 03:26 PM
Lord Voldemort
Re: Limit of a term involvin sum
That's pretty, I really do love Taylor series. Thanks for the response!
• Aug 24th 2013, 04:09 PM
MathoMan
Re: Limit of a term involvin sum
THANKS... twizter, if you ever plan on coming down to visit Croatia, let me know. I owe you a beer, scratch that, I owe you dinner and a beer! ;) )! Thanks both of you for your replies.
• Aug 24th 2013, 04:16 PM
twizter
Re: Limit of a term involvin sum
You are welcome, Mathoman. I have no plans of visiting Croatia anytime soon but I will keep that in the back of my mind (Beer) (Wink)
• Aug 24th 2013, 04:22 PM
MathoMan
Re: Limit of a term involvin sum
Lord Voldemort you're in the round too! :)
• Aug 24th 2013, 04:29 PM
MathoMan
Re: Limit of a term involvin sum
Quote:

Originally Posted by twizter
You are welcome, Mathoman. I have no plans of visiting Croatia anytime soon but I will keep that in the back of my mind (Beer) (Wink)

My old man used to say: "Go treat yourself to a beer. Send me the bill." :D
• Aug 24th 2013, 10:23 PM
Idea
Re: Limit of a term involvin sum
Idea: write down a Riemann sum for the following

$\displaystyle \int _0^1x e^{x^2}dx$