24 - Series

Partial Sum of Series ${\sum }_{k=1}^{\infty }{a}_k$

Let $(a_k)$ be a sequence
For $n\ge 1$, let

$$s_n=a_1+a_2+\cdots +a_n=\sum _{k=1}^n{a}_k$$

Convergence of a Partial Sum

$$\text{Series }\sum _{k=1}^{\infty }\text{ converges if sequence }(s_n)\text{ of partial sum converges}$$

In other words

$$\text{if }{s}_n\to s\text{ as }n\to \infty \text{ then }\sum _{k=1}^{\infty }{a}_k=s$$

If $(s_n)$ does not converge, then

$$\sum _{k=1}^{\infty }{a}_k\text{ diverges }$$

Examples of Series

$$e=\sum _{n=0}^{\infty }\frac{1}{n!}$$ $$e^x=\sum _{n=0}^{\infty }\frac{x^n}{n!}$$

Convergence of Series is a special case of convergence for sequences

Notation on Series

$$\sum _{k\ge 1}{a}_k=\sum a_k=\sum _{k=1}^{\infty }{a}_k$$

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Relation between series and sequences

Consider the series

$$\sum _{k=1}^{\infty }{a}_k$$

If ${\sum }_{k=1}^{\infty }$ converges by assumption then

$$a_k\to 0\text{ as }k\to \infty$$

Tip for showing that a series diverges

If you can show that

$$a_k\nrightarrow 0\text{ as }k\to \infty$$

Then the series $s_n$ doesn’t converge

Proof

Let $s_n={\sum }_{k=1}^n{a}_k$, then $(s_n)$ converges by assumption

Suppose $s_n\to s$ as $n\to \infty$ then

$$s_{n-1}\to s\text{ as }n\to \infty$$

So by AOL

$$a_n=s_n-s_{n-1}\to s-s=0\text{ as }n\to \infty$$

Positive sequence with bounded series converges

Let $(a_k)$ be a sequence of non-negative real numbers
Let $s_k={\sum }_{k=1}^n{a}_k$

Suppose that $(s_n)$ is bounded then

$$\sum _{k=1}^{\infty }{a}_k\text{ converges}$$

Proof

Since $a_k\ge 0$ for all $k$ then $(s_n)$ is increasing
So $(s_n)$ is monotone and bounded, so by Monotone Sequences Theorem

$$(s_n)\text{ converges hence }\sum _{k=1}^{\infty }{a}_k\text{ converges}$$

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Convergence of ${\sum }_{k=1}^{\infty }{k}^{-p}$ lemma

Take $p\in \mathbb{R}$

If $p\le 1$

$$\sum _{k=1}^{\infty }{k}^{-p}\text{ diverges}$$

If $p>1$

$$\sum _{k=1}^{\infty }{k}^{-p}\text{ converges}$$

Proof

1. Case $p\le 0$: $k^{-p}\nrightarrow 0$ as $k\to \infty$ so the series does not converge
2. Case $p=1$: Harmonic Series so doesn’t converge
3. Case $0<p<1$:
   As $k^{-p}>k^{-1}>0$ then by Comparison Test then $\sum k^{-p}$ diverges
   (As $\sum k^{-1}$ diverges - harmonic series)
4. Case $p\ge 2$
   We know that $\sum \frac{1}{k^2}$ converges and as

$$0\le k^{-p}\le k^{-2}$$

Then by the comparison test then $\sum k^{-p}$ converges
5) Case $1<p<2$: TODO LATER =_=

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Rearrangement of Series

Let $g:\mathbb{N}\to \mathbb{N}$ be a bijection
Given a series $\sum a_k$
Let $b_k=a_{g(k)}$
Then

$$\sum b_k\text{ is a rearrangement of }\sum a_k$$

Note that if $\sum a_k$ is absolutely convergent then any rearrangement yields the same limit