Does the divergence of one partial sum imply overall divergence?

Question

Consider a sequence defined by

$$a_{3n-2}=-\frac{1}{2n-1}, a_{3n-1} = \frac{1}{4n-1}, a_{3n} = \frac{1}{4n}$$

I want to show $\sum a_n$ diverges. Is it enough the show that the partial sum $S_{3k} = \sum_{n=1}^k \frac{8k^2-8k+1}{4k(2k-1)(4k-1)}> \sum_{n=1}^k \frac{1}{8k}$ for $k>1$ diverges?

Answer 1

Many confusions in the answers. Just recall the definition:

The nature of a series $\sum a_n$ is the nature of its partial sum $(S_n)_{n\geq 0}$ where $S_n = \sum_{k=0}^na_k$.

So, strictly speaking, showing that $(S_n)_{n\geq 0}$ diverges is showing that $\sum a_n$ diverges.

That being said, the actual problem you have here is knowing if showing that $(S_{3n})_{n\geq 0}$ is enough to show that $(S_n)_{n\geq 0}$ diverges as well. Remember that a divergent sequence is a sequence that does not converge, this includes:

• Sequences that do not have a limit (finite or infinite)

• Sequences that have an infinite limit ($\pm \infty$)

Then, what you would like to use is the following property:

Any sequence with a divergent subsequence is divergent.

Proof: Let $(u_{\phi(n)})_{n\geq 0}$ be a divergent subsequence of $(u_n)_{n\geq 0}$. Then: $$ \forall l > 0, \exists \epsilon>0, \forall N\in \mathbb{N},\exists n \in \mathbb{N}, | u_{\phi(n)}-l | > \epsilon $$ Since $\phi: \mathbb{N} \to \mathbb{N}$, there exists $m \in \mathbb{N}$ so that $\phi(n) = m$. So: $$ \forall l > 0, \exists \epsilon>0, \forall N\in \mathbb{N},\exists m \in \mathbb{N}, | u_m-l | > \epsilon $$ meaning $(u_n)_{n\geq 0}$ is divergent.