Show that $\sum \frac{1}{2k}$ and $\sum \frac{1}{2k+1}$ diverge formality correction 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.  
 A: 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.
A: 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?
A: 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 $1/2$.
