Find all possible real numbers such that the series is convergent.

1. Feb 20, 2014

anthony.g2013

1. The problem statement, all variables and given/known data

Is there a real number c such that the series:

∑ (e - (1+ 1/n)^n + c/n), where the series goes from n=1 to n=∞, is convergent?

3. The attempt at a solution

I used the ratio test by separating each term of the function as usual to find a radius of convergence, but that doesn't seem to work. One can observe though that limit as n→∞, (1 + 1/n)^n = e. But don't know how I can use that in solving this question.

Thanks.

Last edited: Feb 20, 2014
2. Feb 20, 2014

gopher_p

So ratio test didn't work, and it's pretty clear that the test for divergence/nth-term test is gonna be inconclusive. The integral test is gonna be darn near impossible with that $(1+1/x)^x$ term. And the root test probably isn't gonna be very pleasant either. This isn't an alternating series, so that test doesn't even apply.

So what else do you have in your tool chest for determining the convergence/divergence of this series?

3. Feb 20, 2014

anthony.g2013

Can't really think of anything else. Any suggestions?

4. Feb 20, 2014

gopher_p

Do you know any other tests besides the ones that I named?

5. Feb 20, 2014

Dick

I would try to get a series expansion of (1+1/n)^n in powers of (1/n). That will let you figure out what c must be. Then try a limit comparison test to show it converges.

6. Feb 20, 2014

anthony.g2013

Nope. Why don't you suggest one and I can may be look it up?

7. Feb 20, 2014

gopher_p

Dick suggested the limit comparison test. I think that's a good place to start.

8. Feb 20, 2014

Dick

Start with the series expansion idea. I think that's the only way you can figure out what to compare with and prove it.

9. Feb 21, 2014

anthony.g2013

Okay so is it true that lim (1+1/n)^n as n→∞ is the same as ∑ (1+1/n)^n as n goes from 1 to infinity? If so, then

∑ e -e +c/n leaves us with the harmonic series c∑1/n which clearly diverges. So c=0. Please let me know if I am correct. If not, then you might as well give me an answer as all my efforts have failed.

Thanks!

10. Feb 21, 2014

Ray Vickson

No, if all your efforts have failed we should not give you the answer (at least not before the due-date of the assignment) because that would allow you to get marks for somebody else's work.

However, I will give you a bit more of a hint: the important thing to determine is the behaviour of $e - (1 + 1/n)^n$ for large $n$. The best way to figure this out is to look at $L_n = \ln (1 + 1/n)^n = n \ln (1 + 1/n)$ for large n. Of course, its large-$n$ limit is 1 (so you get $e$ when you exponentiate it), but it is important to know as well just how rapidly it approaches 1, so you can assess how rapidly $(1 + 1/n)^n$ approaches $e$. For example, is the approach to $e$ rapid enough that the series would converge if you had c = 0?

11. Feb 21, 2014

Dick

That's not a good enough estimate of the behavior of (1+1/n)^n. I'll give you a hint on how to proceed. You want an expansion in powers of 1/n. So put x=1/n. That turns (1+1/n)^n into (1+x)^(1/x). You want a power series expansion of that. Second hint. That's the same as exp(log(1+x)*(1/x)). Give me the first few terms in the power series expansion of that.

12. Feb 21, 2014

anthony.g2013

The first few terms of power expansion of (1+1/n)^n would be e^(n(1/n -1/2n^2 + 1/3n^3-...))= e^(1- 1/2n +O(n^2)). Now the e terms will cancel and we will be left with e^(-1/2n+O(n^2)). How to proceed form here?

13. Feb 21, 2014

Dick

Now use e^x=1+x+x^2/2+... Be careful with that cancelling e thing. It does more than just cancel the initial e. See how this is going to work?

Last edited: Feb 21, 2014