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I will now show that $\sum \frac{1}{2k}$ and $\frac{1}{2k+1}$ both diverge.

$\exists \ \epsilon > 0 \ \forall N \in \mathbb{N}$ so that for $m>n>N$:

$0\le |\sum_{k=n}^{m} \frac{1}{2k+1} | \le |\sum_{k=n}^{m} \frac{1}{2k}| < |\sum_{k=n}^{m} \frac{1}{k}| > \epsilon $ for some $n>N$

Tell me if this is formally correct. Please.

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    @VVV No, that is not correct. For instance, consider 0<\sum \frac 1 {k^2}<\sum \frac 1 k, where $\sum \frac 1 {k^2}$ converges but $\sum \frac 1 k$ does not. Take a look at AMPerrine's method again.2011-10-28

3 Answers 3

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This isn't formally correct as AMPerrine pointed out. Since $\displaystyle \sum \frac{1}{k}$ diverges you know (since the Harmonic Series diverges) you can choose $N$ large enough so that $\displaystyle \sum_{k=1}^{N}>2M$ for any $M\in\mathbb{R}$. Thus, $\displaystyle \sum_{k=1}^{N}\frac{1}{2k}>M$. Etc. I hope that helps, although maybe assuming that the Harmonic Series diverges is too strong since this is practically equivalent.

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For the first, what would it mean if $\sum\frac{1}{2k}=\frac{1}{2}\left(\sum\frac{1}{k}\right)$ did converge?

For the second, notice that $\frac{1}{2k+1}>\frac{1}{3k}$ for all $k>1$. Does $\sum\frac{1}{3k}$ converge?

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    @VVV AMPerrine is suggesting a proof by contradiction. Assume $\sum \frac 1 {2k}$ converges. Now use his first equality.2011-10-28
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You can use the old proof that the harmonic sum diverges by considering, for any n, $\sum_{k=n+1}^{2n} \frac{1}{k} \ge \sum_{k=n+1}^{2n} \frac{1}{2n} = \frac{n}{2n} = \frac{1}{2}. $

Therefore, for any $n$ there is an $m$ (in particular, $2n$) such that the sum of n terms and the sum of m terms differ by at least $1/2$.

You can do the same thing for the sums of $1/(2n)$ and $1/(2n+1)$, and you will get $1/4$ instead of $1/2$.