30. Series - more results and another test for convergence

Cauchy Convergence Criterion for Series

Let $(a_k)$ be a sequence, and write

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

Then ${\sum }_{k=1}^{\infty }{a}_k$ converges if and only if

$$\forall ϵ>0\exists N\in \mathbb{N}\text{ such that }\forall n>m\ge N:|s_n-s_m|=\left|\sum _{k=m+1}^n{a}_k\right|<ϵ$$

Proof

Follows from Caughy Convergence Criterion

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Absolute Convergence

Let $(a_k)$ be a sequence then

$$\sum _{k=1}^{\infty }{a}_k\text{ converges absolutely}\text{ if }\sum _{k=1}^{\infty }|a_k|\text{ converges}$$

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Absolute Convergence implies Convergence

Let $(a_k)$ be a sequence

If ${\sum }_{k=1}^{\infty }|a_k|$ converges then

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

Proof

Let

$$s_n=\sum _{k=1}^n{a}_k\text{ and }{S}_n=\sum _{k=1}^n|a_k|$$

For $n>m$ then

$$|s_n-s_m|=\left|\sum _{k=m+1}^n{a}_k\right|\le \sum _{k=m+1}^n|a_k|=|S_n-S_m|$$

As $\sum |a_k|$ converges by assumption
Then $(S_n)$ is Cauchy by Cauchy Convergence Criterion
So $(s_n)$ is Cauchy by the inequality above
Hence $\sum a_k$ converges by the Cauchy Convergence Criterion

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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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Alternating Series Test

Let $(u_k)$ be a real sequence, and consider the series

$$\sum _{k=1}^{\infty }(-1{)}^{k-1}{u}_k$$

If

1. $u_k\ge 0$ for $k\ge 1$
2. $(u_k)$ is decreasing, that is $u_{k+1}\le u_k$ for $k\ge 1$
3. $u_k\to 0$ as $k\to \infty$
   Then

$$\sum _{k=1}^{\infty }(-1{)}^{k-1}{u}_k\text{ converges}$$

Proof

Let $s_n={\sum }_{k=1}^n(-1{)}^{k-1}{u}_k$

1. $(s_{2n})$ is bounded above as

$$s_{2n}=u_1-(u_2-u_3)-(u_4-u_5)-\cdots -(u_{2n-2}-u_{2n-1})-u_{2n}\le u_1$$

So $u_1$ is an upper bound of $(s_{2n})$
2) $(s_{2n})$ is increasing as

$$s_{2n+2}-s_{2n}=u_{2n+1}-u_{2n+2}\ge 0$$

So by the Monotone Sequences Theorem, $(s_{2n})$ converges
Suppose $s_{2n}\to s$ as $n\to \infty$
As

$$s_{2n+1}=s_{2n}+u_{2n+1}\to s+0=s\text{ as }n\to \infty \text{ by AOL}$$

So $(s_{2n+1})$ also converges to $s$

There is $N_1\in \mathbb{N}$ such that for $n\ge N_1$ then

$$|s_{2n}-s|<ϵ$$

There is $N_2\in \mathbb{N}$ such that for $n\ge N_2$ then

$$|s_{2n+1}-s|<ϵ$$

Let $N=\max \{2N_1,2N_2+1\}$ then for $n\ge N$ then

1. If $n$ is even then $n=2k$ for some $k$ so $|s_n-s|=|s_{2k}-s|<ϵ$
2. If $n$ is odd then $n=2k+1$ for some $k$ so $|s_n-s|=|s_{2k+1}-s|<ϵ$

Hence $|s_n-s|<ϵ$ so $(s_n)$ converges

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