Question for math purists: is the following proof related to harmonic series wrong? I am reading a proof of the fact that a harmonic series does not converge. It seems to me that the proof (of course not the fact itself) is wrong, is it?

We assume that a harmonic series converges to $s \in \mathbb R$ (a
  proof by contradiction).
For an integer $m > 1$
  $$\sum_{j=1}^{\infty}\frac{1}{j}=\sum_{j=1}^{m}\frac{1}{j}+\sum_{j=m+1}^{2m}\frac{1}{j}+\sum_{j=2m+1}^{3m}\frac{1}{j}+\cdots$$

My explanation of the statement: it is true because $\sum_{j=1}^{\infty}\frac{1}{j}=\lim_{n \to \infty} \sum_{j=1}^{n}\frac{1}{j} = \lim_{n \to \infty} s_n$
$$\sum_{j=1}^{m}\frac{1}{j}+\sum_{j=m+1}^{2m}\frac{1} {j}+\sum_{j=2m+1}^{3m}\frac{1}{j}+\cdots=\sum_{k=1}^{\infty}\sum_{j=(k-1)m+1}^{mk}\frac{1}{j}=\lim_{n \to \infty}\sum_{k=1}^{n}\sum_{j=(k-1)m+1}^{mk}\frac{1}{j}=\lim_{n \to \infty}s'_n$$
Because $s'_n$ is a sub-sequence of $s_n$ and $s_n \to s \implies s'_n \to s$, thus the equality above.
Now the proof proceeds by considering another series $s''_n=\sum_{k=1}^{n}\sum_{j=(k-1)m+1}^{mk}\frac{1}{mk}$. It is clear that $s'_n > s''_n$ $\forall n$, so the author concludes that $\lim_{n \to \infty}s'_n=s>\lim_{n \to \infty}s''_n$ which he presents in the following form as something fully obvious: 

$$\sum_{j=1}^\infty \frac{1}{j}>m\left(\frac{1}{m}\right)+m\left(\frac{1}{2m}\right)+m\left(\frac{1}{3m}\right)+\cdots$$

The author finishes the proof by noting that because $s''_n$ is itself a harmonic series we get that $s>s$ (contradiction).
What seems to me wrong is the conclusion $\sum_{j=1}^\infty \frac{1}{j}>m(\frac{1}{m})+m(\frac{1}{2m})+m(\frac{1}{3m})+\cdots$, i.e. $\lim_{n \to \infty}s'_n>\lim_{n \to \infty}s''_n$ because $s'_n > s''_n$ $\forall n$. Does it seem wrong to you?
 A: Certainly, it's not automatically the case that if $(a_n)$ and $(b_n)$ are convergent sequences and $a_n>b_n$ for all $n$, that then $\lim_{n\to\infty}a_n>\lim_{n\to\infty}b_n$.  Instead, the most you can say is that $\lim_{n\to\infty}a_n\ge\lim_{n\to\infty}b_n$.  

Example: Let $a_n$ be the sequence >$$0,\frac12,\frac23,\frac34,\frac45\dots$$, and let $b_n$ be the same sequence, but lagging behind by one:
  $$0,0,\frac12,\frac23,\frac34,\dots$$

Then $a_n>b_n$ for all $n$, but both sequences converge to the same limit.  
However, if we can bound the difference $a_n-b_n$ below by some constant $\delta>0$, then we must have actual inequality $\lim_{n\to\infty}a_n>\lim_{n\to\infty}b_n$  Exercise: prove this.
And that is exactly what we have in this case.  Note that 
\begin{align}
s_n'=\left(\frac11+\dots+\frac1m\right)+&\dots+\left(\frac1{(n-1)m+1}+\dots+\frac{1}{nm}\right)\\
s_n''=\left(\frac1m+\dots+\frac1m\right)+&\dots+\left(\frac1{nm}+\dots+\frac{1}{nm}\right)\\
\end{align}
Then notice that each (bracketed) term of the sum $s_n''$ is greater than the corresponding term of the sum $s_n'$.  This means that we can write
$$
s_n'-s_n''>\left(\frac11+\dots+\frac1m\right)-\left(\frac1m+\dots+\frac1m\right)>0
$$
So not only do we know that $s_n'>s_n''$ for all $n$, we can bound the difference below.  Therefore, the limit of the $s_n'$, if it exists, must be strictly greater than the limit of the $s_n''$. 
