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.
Ok guys, it seems that I have become rusty since I cant find an apropriate way to tackle the following limit:
.
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...
Now it seems one can conlude that
.
That means that .
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.
Thanks in advance, MathoMan.
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.
I was hoping someone could enlighten us, but I'll show you how I arrived to a better upper bound of e:
Let ; note that so we have and thus so we can say:
which has a nice representation (see the second one: List of mathematical series - Wikipedia, the free encyclopedia ). Thus:
The approach I tried before that led to as an upper bound was similar except I used the fact that
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.
Here's how I did it. First pull out the e from outside of the summation. Now expand using the Taylor series expansion for e^x. Now multiply this by . You get . Now consider . The leading term of this is . Now this is the only term that matters in the limit since it is being divided by . So it reduces to . Factoring out the 1/2 from this we get which is nothing but . This sum is equal to e -1. So the answer is . Hope this helps