series

Series

Let $(a_n{)}_{n\in \mathbb{N}}$ be a sequence. The $n$-th partial sum is $s_n:={\sum }_{k=1}^n{a}_k$.
The series ${\sum }_{k=1}^{\infty }{a}_k$ is the limit of the partial sums:

$$\sum _{k=1}^{\infty }{a}_k:=\underset{n\to \infty }{\lim }{s}_n=\underset{n\to \infty }{\lim }\sum _{k=1}^n{a}_k$$

If this limit exists (finite), the series converges and we call the sequence $(a_n)$ summable.
If $s_n\to \pm \infty$, the series is definitely divergent.
Otherwise (partial sums keep bouncing, e.g. $\sum (-1{)}^k$) the series is divergent.

A series is just a sequence of partial sums. All the tools for sequences (convergence, boundedness, Cauchy, …) apply to $(s_n)$.

Every summable sequence is a null sequence

$$\sum _{k=1}^{\infty }{a}_k\text{is convergent}⟹(a_n)\to 0$$

If $(a_n)\nrightarrow 0$, the series diverges.
It’s a necessary but insufficient condition: $a_n=1/n\to 0$ but $\sum 1/k=\infty$ (see harmonic series).

Proof

Cauchy series definition:

$$\forall ϵ>0\exists n_0\in \mathbb{N}\forall m>n\ge n_0:|\sum _{k=n+1}^m{a}_k|<ϵ$$

Set $m=n+1$ to get:

$$\forall ϵ>0\exists n_0\in \mathbb{N}\forall n\ge n_0:|a_{n+1}|<ϵ$$

which is exactly the definition of $a_n\to 0$.

What needs to be true about $(a_n)$ for ${\sum }_{k=1}^{\infty }{a}_k$ to be convergent?

It needs to be a null sequence.

${\sum }_{k=0}^{\infty }{q}^k$ can only be convergent for $|q|<1$

Calculation rules

If $\sum a_k$ and $\sum b_k$ both converge, then:

$$\sum (a_k+b_k)=\sum a_k+\sum b_k,\sum c\cdot a_k=c\cdot \sum a_k$$

Non-negative sequences of partial sums converge iff they are bounded

If $a_k$ is a non-negative sequence, i.e. $a_k\ge 0$ for all $k$, then the series $s_n:={\sum }_{k=1}^n{a}_k$ is non-decreasing, then, by the monotonicity principle, iff the series ${\sum }_{k=1}^{\infty }$ converges, it is bounded, i.e.,

$$\exists C\in \mathbb{R}\forall n\in \mathbb{N}:s_n\le C$$

See also: geometric series, harmonic series

Are there any non-negative sequences of partial sums that are unbounded but converge?

No. They converge iff they are bounded.