How to prove rigorously that the series $1/(3n)$ diverges How can I prove rigorously that the series
$$\sum_{n=1}^\infty \frac{1}{3n}$$ 
Diverges, assuming that I know that the harmonic series when $p = 1$ diverges,
I thought of using the property
$$\sum_{n=1}^\infty ca_n = c\sum_{n=1}^\infty a_n$$ 
However I think this only works when both of the series converges?
So to summarize, how can I prove that the series diverges knowing that the p-series $$\sum_{n=1}^\infty \frac{1}{n}$$ diverges? 
 A: Proof by contradiction. Let us assume that the series
$$
  \sum_{n=1}^\infty \frac{1}{3n}
$$
converges to, say, $A$. Then it would be also true that
$$
  A = \frac{1}{3} \sum_{n=1}^\infty \frac{1}{n}
$$
which implies that
$$
  \sum_{n=1}^\infty \frac{1}{n}
$$
converges to $3A$. Knowing the fact that this series diverges (we found a contradiction) completes the proof by contradiction.
A: You do not need convergence, just consider partial sum, you can factorize the scalar because you manipulate a finite series:
$$\sum_\limits{n=1}^N \dfrac 1{3n}=\dfrac 13\underbrace{\sum_\limits{n=1}^N \dfrac 1n}_{\to+\infty}\to+\infty$$
Thus you get that the partial sum does not have a finite limit so the series diverges.
A: Yes, it's true you use that property, but it's good you noticed that this equality is only guaranteed if the series converges. Actually, let's write the statement more precisely:

If $\sum_{n=1}^\infty a_n$ converges, then so does $\sum_{n=1}^\infty c\cdot a_n$ for any constant $c$. 

Now think on the contrapositive of this statement:

If $\sum_{n=1}^\infty c\cdot a_n$ does not converge for some constant $c$, then $\sum_{n=1}^\infty a_n$ does not converge.

Do you see how you could use this to prove our statement?
There is a hint below.

 Hint: Take $a_n = 1/(3n)$

A: Yeah, you can use a variation of your property. For any constant $c \neq 0$, we have $\sum_{i=1}^{\infty} ca_{n}$ converges if and only if $\sum_{i=1}^{\infty} a_{n}$ converges and likewise with divergence.
Specifically for this problem, assume your series converges. Then we have that, by definition, there exists some $L$ such that for any $\epsilon > 0$, there exists an $N \in \mathbb{N}$ such that for $n \geq N$, we have
$$\left| \sum_{i=1}^{n} \frac{1}{3n} - L \right| < \frac{1}{3}{\epsilon}.$$
But this is if and only if
$$\left| \sum_{i=1}^{n} \frac{1}{n} - 3L \right| < \epsilon$$
and since $\epsilon$ was made arbitrary, this implies that $\sum_{i=1}^{\infty} \frac{1}{n}$ converges to $3L$, and you know that it doesn't. There's not much else to generalizing this concept for any series.
A: You can use the same proof that $\sum \frac 1n$ diverges. i.e. $1 + \frac 12 + (\frac 13 +\frac 14) + (\frac 15 + \cdots \frac 18) + (\frac 19 + \cdots + \frac 1{16})+\cdots < 1 + \frac 12 +\frac 12 + \frac 12 +\cdots$
And a divergent series multiplied by a constant (other than 0), indeed produces divergent series.
A: Note that by definition
$S_n(k)=\sum_{n=1}^k \frac{1}{n}$ diverges $\iff \forall M>0 \quad \exists \bar k$ such that $S_n(k)>M \quad \forall k>\bar k$
then also
$S_{3n}(k)=\frac13S_n(k)=\sum_{n=1}^k \frac{1}{3n}$ diverges since
$\forall M>0 \quad \exists \bar k$ such that $S_n(k)>3M \quad \forall k>\bar k\implies S_{3n}(k)>M$
