Is this series bounded?

  • Thread starter emptyboat
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  • #1
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Main Question or Discussion Point

I asked by someone.
But I can't answer to it.
[tex]\sum^{\infty}_{1}(-1)^n*(1+\frac{1}{n})^n[/tex]
Is this series bounded?
I can't do anything about that.
 
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Answers and Replies

  • #2
jbunniii
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Do you know what

[tex]\lim_{n \rightarrow \infty} \left(1 + \frac{1}{n}\right)^n[/tex]

equals?
 
  • #3
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I know this series diverge by lim an=e, not zero.
But I can't figure out it's bounded.
Would you give me more informations?
 
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  • #4
907
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The sum of any two consecutive terms goes to zero as O(1/n^2). Since the sum of 1/n^2 is convergent, you can prove that your original series is bounded.
 
  • #5
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hmm... Do you mean (1 + 1/(n+1))^(n+1) - (1 + 1/n)^n ≤1/n^2?
I can not understand it. Give me more information, plz.
 
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  • #6
907
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hmm... Do you mean (1 + 1/(n+1))^(n+1) - (1 + 1/n)^n ≤1/n^2?
I can not understand it. Give me more information, plz.
Correct. You need to prove that there's such N and C that, for all n>N,

(1 + 1/(n+1))^(n+1) - (1 + 1/n)^n ≤ C/n^2

You can do that if you write

(1+1/n)^n = exp(n log (1+1/n))

and then use Taylor expansion of log.
 
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  • #7
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I computed the process you have given.
log(x)=(x-1) - 1/2*(x-1)^2 + 1/3*(x-1)^3 - ...(correct?)
So My result is
exp (1 - 1/2(n+1) + 1/3(n+1)^2 - ...)
- exp (1 - 1/2n + 1/3n^2 - ...)
let exp(A(n+1)) - exp((An))
exp does not linear, I can't do further.
Help me more...
 
  • #8
907
2
[tex]exp(1 - \frac{1}{2n} + \frac{1}{3n^2} + ...) \approx e*( - \frac{1}{2n} + \frac{1}{3n^2}) + ( - \frac{1}{2n} + \frac{1}{3n^2})^2/2 + ...) = e*(-\frac{1}{2n}+\frac{1}{3n^2} + \frac{1}{8n^2} + ...) = e*(-\frac{1}{2n}+\frac{11}{24n^2} + ... )[/tex]

where ... are terms that go down as 1/n^3 or faster.

[tex]|\frac{1}{2(n+1)} - \frac{1}{2n}| = \frac{1}{2n(n+1)} < \frac{1}{2n^2}[/tex]
 
  • #9
21
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Oh I see!!!
Thank you for your advice.
I think this problem is very hard to solve!!!
 

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