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

[anthony.g2013]

Homework Statement

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?

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.

 

[gopher_p]

Homework Statement

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?

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 not use how I can use that in solving this question.

Thanks.

So ratio test didn't work, and it's pretty clear that the test for divergence/nth-term test is going to be inconclusive. The integral test is going to be darn near impossible with that $(1+1/x)^x$ term. And the root test probably isn't going to 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?

 

[anthony.g2013]

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

 

[gopher_p]

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

 

[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.

 

[anthony.g2013]

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

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

 

[gopher_p]

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

 

[Dick]

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

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

 

[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!

 

[Ray Vickson]

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!

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?

 

[Dick]

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!

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.

 

[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?

 

[Dick]

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?

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?