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?
Best Answer
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.