Explain why ∑(1+n)/(1+2n) is divergent

  • Thread starter Jamin2112
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  • #1
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Homework Statement



As the title says.

Homework Equations



mentioned in solution

The Attempt at a Solution



Let Sn = {(1+1)/(1+2) , (1+2)/(1+4), (1+3)/(1+6), ...}. If ∑(1+n)/(1+2n) is convergent, then lim n-->∞ Sn = 0; to put it another way, there exists an N so that whenever n ≤ N,

|(1+n)/(1+2n)|=(1+n)/(1+2n) < ∂ for all ∂ > 0.

(1+n)/(1+2n) < ∂ ----> (1+2n)/(1+n) > 1/∂ ----> 1 + n/(n+1) > 1/∂.

But since n/(n+1) < 1 for all n, the inequality 1 + n/(n+1) < 1/∂ fails when ∂ ≤ 1/2.

Thus the series converges.
 

Answers and Replies

  • #2
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But we have

[tex]\lim_{n\rightarrow +\infty}{\frac{1+n}{1+2n}}=\frac{1}{2}[/tex]

So the series diverges...
 
  • #3
986
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But we have

[tex]\lim_{n\rightarrow +\infty}{\frac{1+n}{1+2n}}=\frac{1}{2}[/tex]

So the series diverges...

Huh?
 
  • #4
Dick
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Huh?
Why "huh?". The limit of the terms in your series isn't zero. Are you trying to derive a contradiction by assuming it is? If so, that's an odd way to go about proving the series doesn't converge.
 
  • #5
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Why "huh?". The limit of the terms in your series isn't zero. Are you trying to derive a contradiction by assuming it is? If so, that's an odd way to go about proving the series doesn't converge.
I thought I did find a contradiction. I assumed it was convergent, then I used the definition of convergence and found a flaw (the "for all epsilon").
 
  • #6
Dick
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I thought I did find a contradiction. I assumed it was convergent, then I used the definition of convergence and found a flaw (the "for all epsilon").
I suppose you could do it that way. It's just awkward. Most people would just compute the limit of the Sn, conclude it's 1/2 and then say the series diverges because 1/2 isn't 0. I don't think you need an epsilon-delta proof here.
 

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