Proving series is divergent by definition

Question

I was trying to prove the divergence of the series $\sum_{n=1}^{\infty} a_n$

when $a_n$ is defined by $a_n = \begin{cases} \frac{1}{n} & n\enspace is \enspace odd\\ \frac{1}{n^3} & n\enspace is \enspace even \end{cases}$
I tried to do the following but I feel like I am missing something.
Let us denote the (2n-1)th partial sum of $a_n$ by $S_{2n-1} = \sum_{k=1}^{2n-1}a_n = 1 + \frac{1}{2^3} + \frac{1}{3} + \frac{1}{4^3} + ... + \frac{1}{2n-1}$ therefore :
$S_{2n-1} \geq 1 + \frac{1}{3} + \frac{1}{5} + ... + \frac{1}{2n-1} = T_n = \sum_{k=1}^n \frac{1}{2n-1} $
$T_n$ is the n-th partial sum of $\sum \frac{1}{2n-1}$ which is divergent therefore $\lim_{n \to \infty} T_n = \infty$ .
Therefore $\lim_{n \to \infty}S_{2n-1} = \infty$ and therefore the series is divergent.
My problem is that I am not sure about the last step which and if whether or not $\lim_{n \to \infty}S_{2n-1} = \infty$ implies that the series is divergent, do I have also to consider the sequence $S_{2n}$ in order to determine that the series is divergent?
Thanks in advance.

Answer 1

If you want to prove that $\lim S_n = +\infty$ then you must consider $S_{2n}$, writing for example $S_{2n}>S_{2n-1}$ combined with $S_{2n-1} \to +\infty$ is sufficient.

If you just want to prove that $S_n$ is not convergent in $\mathbb{R}$, then writing that there is one subsequence that is not convergent is sufficient.

Answer 2

Shortcut

All of the terms are positive.

Ignore the even terms. The odd terms, by themselves are strictly greater than $\displaystyle \frac{1}{2} \times ~\sum_{i=1}^\infty \frac{1}{n}~$ (i.e. the harmonic series), which is known to be divergent (i.e. goes to $\infty$).

$\frac{1}{2} \times \infty = ~$ (informally) $~\infty$.

That is, since your series is greater than (1/2) of a divergent series (i.e. a series that goes to $\infty$), your series must also go to $\infty$ (i.e. must also be divergent).

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See also the comments following this answer.

Answer 3

Proof by making contradictions: Assume $\sum_{n=1}^\infty a_n $ is convergent.

Since all terms are positive, it is absolutely convergent, so no worries about rearrangement.

$$\sum_{n\in odd}^\infty a_n=\sum_{n=1}^\infty a_n -\sum_{n\in even}^\infty a_n$$

Based on our assumption, the RHS is convergent, so the LHS is convergent. But we know the LHS is actually divergent due to p-series test. Therefore, we get a contradiction.