The series $\sum_{n=1}^\infty\frac1n$ diverges! We all know that the following harmonic series 
$$\sum_{n=1}^\infty\frac1n=\frac 1 1 + \frac 12 + \frac 13 + \cdots $$ 
diverges and grows very slowly!! I have seen many proofs of the result but recently found the following: $$S =\frac 1 1 + \frac 12 + \frac 13 +\frac 14+ \frac 15+ \frac 16+ \cdots$$ $$> \frac 12+\frac 12+ \frac 14+ \frac 14+ \frac 16+ \frac 16+ \cdots =\frac 1 1 + \frac 12 + \frac 13 +\cdots = S.$$
In this way we see that $S > S$. 
Can we conclude from this that $S$ is divergent??
 A: If $\sum_{n=1}^\infty a_n$ and $\sum_{n=1}^\infty b_n$ both exist, $a_n\ge b_n$ for all $n$, and $a_i>b_i$ for at least one $i$, then the first sum must be strictly greater than the second. This is because the first's partial sum is eventually always at least $a_i-b_i$ more than the second's partial sums. In this case, one can subsequently reason that if the first exists, so does the second. If this is your reasoning, it is valid.
A: The proof can be made a bit more rigorous by setting
$$
\begin{align}
a_n=\frac1n:&\,\quad1,\,\frac12,\frac13,\frac14,\frac15,\frac16,\dots\\b_n=\frac1{2\lfloor(n+1)/2\rfloor}:&\quad\frac12,\frac12,\frac14,\frac14,\frac16,\frac16,\dots
\end{align}\tag{1}
$$
Note that $a_n\ge b_n$, $a_n\gt b_n$ when $n$ is odd, and $a_n=b_{2n-1}+b_{2n}$.
Assuming that
$$
\sum_{n=1}^\infty a_n\tag{2}
$$
converges, then
$$
\sum_{n=1}^\infty b_n=\sum_{n=1}^\infty(b_{2n-1}+b_{2n})=\sum_{n=1}^\infty a_n\tag{3}
$$
also converges. However,
$$
\sum_{n=1}^\infty(a_n-b_n)\gt0\tag{4}
$$
Since $a_n\ge b_n$ and $a_n\gt b_n$ when $n$ is odd.
Now, $(3)$ says that
$$
\sum_{n=1}^\infty b_n=\sum_{n=1}^\infty a_n\tag{5}
$$
and $(4)$ says that
$$
\sum_{n=1}^\infty b_n\lt\sum_{n=1}^\infty a_n\tag{6}
$$
These last two statements are contradictory, so the assumption that $(2)$ converges must be false.
A: There's another way to approach this, via integration: compare - on the domain $(1, \infty)$ the curve $y_1={1\over x}$ with the step function $y_2={1\over Floor(x)}$ (where $Floor(x)$ is the greatest integer $<x$).
Clearly, for every $x\in (1,\infty)$, we have $0<y_1(x)\le y_2(x)$, so $\int_1^\infty y_1dx\le\int_1^\infty y_2dx$; moreover, $\int_1^\infty y_2dx$ is just the sum of the harmonic series.
But integrating, we get $\int_1^\infty y_1dx=\ln(x)\vert^\infty_1=\infty$, so the harmonic series must diverge.

Of course, this is non-rigorous, but it's good motivation, and it can be made rigorous without much work.
A: Here is another approach to making the answer more rigorous. Assume the series converges, then
$$
\begin{align}
\sum_{k=1}^\infty\frac1k
&=\sum_{k=1}^\infty\left(\frac1{2k-1}+\frac1{2k}\right)\\
&=\sum_{k=1}^\infty\left[\left(\frac1{2k}+\frac1{2k}\right)+\left(\frac1{2k-1}-\frac1{2k}\right)\right]\\
&=\sum_{k=1}^\infty\frac1k+\sum_{k=1}^\infty\left(\frac1{2k-1}-\frac1{2k}\right)\\
&=\sum_{k=1}^\infty\frac1k+\log(2)
\end{align}
$$
The rearrangements are justified because the series are all of positive terms.
The sum
$$
\begin{align}
\sum_{k=1}^\infty\left(\frac1{2k-1}-\frac1{2k}\right)
&=\sum_{k=1}^\infty\frac{(-1)^{k-1}}k\\[6pt]
&=\log(2)
\end{align}
$$
is a well-known series for $\log(2)$.
A: The standard proof that everyone has seen but no one  (or at least not i) can remember is to group more and more terms together,  in such a way as to see the series is larger than $1+\frac 12+\frac 12+\dots  $...
Why not include that here ?...
Namely if $S_n $ is the $n $-th partial sum, just note that $S_{2n}-S_n=\frac 1 {n+1}+\dots +\frac1{2n}\gt \frac 12 $...
Thus the sequence of partial sums is not Cauchy, so the series doesn't converge... (or by comparison with $1+\frac 12+\frac 12+\dots  $)
