TITLE: Comparison test of convergence or divergence of Harmonic series
QUESTION [0 upvotes]: Is it possible to show convergence
or divergence of  $\displaystyle \sum^{\infty}_{k=1}\frac{1}{3k}$ by using comparison with $\displaystyle \sum^{\infty}_{k=1}\frac{1}{k}?$

What I tried:
If $\displaystyle \sum^{\infty}_{k=1}\frac{1}{k}$ diverges (using Harmonic series sum) then $\displaystyle \frac{1}{4}\sum^{\infty}_{k=1}\frac{1}{k}$ also diverges.
But I did not understand how I could use comparison test here.
Can somebody help me please?

REPLY [2 votes]: @A learner gave you a solution directly using the comparison test. Another way,

$${\sum_{k=1}^{\infty}\frac{1}{3k}=\lim_{n\to\infty}\sum_{k=1}^{n}\frac{1}{3k}}$$
Looking at the partial sums,
$${\sum_{k=1}^{n}\frac{1}{3k}=\frac{1}{3}\sum_{k=1}^{n}\frac{1}{n}=\frac{H_n}{3}}$$
Where ${H_n}$ is used here to represent the ${n^{th}}$ partial sum of the Harmonic Series. Now just argue that ${H_n}$ increases without bound (it diverges to infinity) and so the partial sums for ${\sum_{k=1}^{n}\frac{1}{3k}=\frac{H_n}{3}}$ must also diverge
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