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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?

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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.

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  • $\begingroup$ Thank you I didn't think about it just began discrete maths so I'm not used to using other forms of proof than the direct one so thank you for the tip!! Will try to consider more proof techniques in the future! $\endgroup$ – Ian Leclaire Jan 27 '18 at 23:35
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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.

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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 $a_n = 1/(3n)$

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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.

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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.

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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 $S_n(k)>3M \quad \forall k>\bar k\implies S_{3n}(k)>M$

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