# Thread: Limit of a term involvin sum

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

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 ) , 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.

2. ## 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.

3. ## 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.

4. ## 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 $\displaystyle \frac{e}{n^2} \sum\limits_{i=1}^{n}i\cdot {\left(e^{\frac{1}{n}}\right)^{i} \le \frac{e}{n} \sum\limits_{i=0}^{n}{\left(e^{\frac{1}{n}}\right) ^{i}$

5. ## 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.

6. ## 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?

7. ## 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

8. ## Re: Limit of a term involvin sum

That's pretty, I really do love Taylor series. Thanks for the response!

9. ## 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.

10. ## 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

11. ## Re: Limit of a term involvin sum

Lord Voldemort you're in the round too!

12. ## Re: Limit of a term involvin sum

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
My old man used to say: "Go treat yourself to a beer. Send me the bill."

13. ## Re: Limit of a term involvin sum

Idea: write down a Riemann sum for the following

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