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

In summary, the conversation discusses the convergence of a series and the use of different tests to determine its behavior. The limit comparison test is suggested, but it is determined that a series expansion of (1+1/n)^n is necessary to accurately compare it to the given series. The conversation then moves on to discussing the power series expansion and how to proceed with using it to solve the problem.
  • #1
anthony.g2013
11
0

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.
 
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  • #2
anthony.g2013 said:

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?
 
  • #3
Can't really think of anything else. Any suggestions?
 
  • #4
Do you know any other tests besides the ones that I named?
 
  • #5
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
gopher_p said:
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?
 
  • #7
Dick suggested the limit comparison test. I think that's a good place to start.
 
  • #8
gopher_p said:
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.
 
  • #9
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
anthony.g2013 said:
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?
 
  • #11
anthony.g2013 said:
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.
 
  • #12
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
anthony.g2013 said:
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?
 
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Related to Find all possible real numbers such that the series is convergent.

1. What is a convergent series?

A convergent series is a sequence of numbers that approaches a definite limit as the number of terms increases. In other words, the sum of the terms in a convergent series will eventually approach a finite value.

2. How do you determine if a series is convergent?

To determine if a series is convergent, you can use various tests such as the ratio test, comparison test, or the integral test. These tests check for certain patterns in the series to determine if it will approach a finite limit or continue to increase without bound.

3. Are there any general rules for finding real numbers that make a series convergent?

There are no specific rules or formulas for finding real numbers that make a series convergent. It depends on the specific series and what test is used to determine convergence. However, some common strategies include using algebraic manipulations, factoring, or simplifying the series.

4. Can a series have more than one set of real numbers that make it convergent?

Yes, a series can have multiple sets of real numbers that make it convergent. This is because different tests may lead to different sets of numbers that satisfy the criteria for convergence. It is important to carefully consider which test to use and to check for any possible alternative solutions.

5. What is the significance of finding all possible real numbers that make a series convergent?

Finding all possible real numbers that make a series convergent can help us better understand the behavior of the series and its sum. It can also provide insights into the properties of the series and how it relates to other mathematical concepts. Additionally, knowing the range of real numbers that make a series convergent can be helpful in making approximations and predictions in real-world applications.

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