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

[Jamin2112]

Homework Statement

As the title says.

Homework Equations

mentioned in solution

The Attempt at a Solution

Let S_n = {(1+1)/(1+2) , (1+2)/(1+4), (1+3)/(1+6), ...}. If ∑(1+n)/(1+2n) is convergent, then lim _n-->∞ S_n = 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.

 

[micromass]

But we have

$$\lim_{n\rightarrow +\infty}{\frac{1+n}{1+2n}}=\frac{1}{2}$$

So the series diverges...

 

[Jamin2112]

But we have

$$\lim_{n\rightarrow +\infty}{\frac{1+n}{1+2n}}=\frac{1}{2}$$

So the series diverges...

Huh?

 

[Dick]

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.

 

[Jamin2112]

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").

 

[Dick]

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.