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

## 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.

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But we have

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

So the series diverges...

But we have

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

So the series diverges...

Huh?

Dick
Homework Helper
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.

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