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Sequences and Series

Sequences​

A sequence is a string of indexed numbers a1,a2,a3,…a_1, a_2, a_3, \ldots. We denote this sequence with (an)n≥1(a_n)_{n\geq1}.

Details​

In a sequence the same number can appear several times in different places.

Examples​

Example

(1n)n≥1 is the sequence 1,12,13,14,…(\displaystyle\frac{1}{n})_{n\geq1} \text{ is the sequence } 1, \displaystyle\frac{1}{2}, \displaystyle\frac{1}{3}, \displaystyle\frac{1}{4}, \ldots

Example

(n)n≥1 is the sequence 1,2,3,4,5,…(n)_{n\geq1} \text{ is the sequence } 1, 2, 3, 4, 5, \ldots

Example

(2nn)n≥1 is the sequence 2,8,24,64,…(2^nn)_{n\geq1} \text{ is the sequence } 2, 8, 24, 64, \ldots

Convergent Sequences​

A sequence ana_n is said to converge to the number bb if for every ε>0\varepsilon >0 we can find an N∈NN\in \mathbb{N} such that ∣an−b∣<ε|a_n-b| < \varepsilon for all n≥Nn \geq N. We denote this with lim⁡n→∞an=b\lim_{n\to\infty}a_n=b or an→ba_n\to b, as n→∞n\to\infty.

Details​

A sequence ana_n is said to converge to the number bb if for every ε>0\varepsilon >0 we can find an N∈NN\in \mathbb{N} such that ∣an−b∣<ε|a_n-b| < \varepsilon for all n≥Nn \geq N. We denote this with lim⁡n→∞an=b\lim_{n\to\infty}a_n=b or an→ba_n\to b, as n→∞n\to\infty.

If xx is a number, then

(1+xn)n→ex as n→∞(1 + \displaystyle\frac{x}{n})^n \rightarrow e^x \text{ as } n\rightarrow\infty

Examples​

Example

The sequence (1n)n≥∞(\displaystyle\frac{1}{n})_{n\geq\infty} converges to 00 as n→∞n\to\infty.

Example

If x is a number, then

(1+xn)n→ex as n→∞(1 + \displaystyle\frac{x}{n})^n \rightarrow e^x \text{ as } n\rightarrow\infty

Infinite Sums (series)​

We are interested in, whether infinite sums of sequences can be defined.

Details​

Consider a sequence of numbers, (an)n→∞(a_n)_{n\to\infty}.

Now define another sequence (sn)n→∞,(s_n)_{n\to\infty}, where

sn=∑k=1naks_n=\displaystyle\sum_{k=1}^na_k

If (sn)n→∞(s_n)_{n\to\infty} is convergent to S=lim⁡n→∞sn,S=\lim_{n\to\infty}s_n, then we write

S=∑n=1∞anS=\displaystyle\sum_{n=1}^{\infty}a_n

Examples​

Example

If

ak=xk,qquadk=0,1,…a_k = x^k, qquad k=0,1,\dots

then

sn=∑k=0nxk=x0+x1+… .+xns_n=\displaystyle\sum_{k=0}^{n}x^k=x^0+x^1+\dots.+x^n

Note also that

xsn=x(x0+x1+… .+xn)=x+x2+⋯+xn+1xs_n=x(x^0+x^1+\dots.+x^n)= x + x^2 + \dots + x^{n+1}

We have

sn=1+x+x2+⋯+xns_n = 1 + x + x^2 + \dots + x^n

xsn=x+x2+⋯+xn+xn+1xs_n = x + x^2 + \dots +x^n + x^{n+1}

sn–xsn=1−xn+1s_n – xs_n = 1 - x^{n+1}

i.e.

sn(1−x)=1−xn+1s_n(1-x) = 1-x^{n+1}

and we have

sn=1−xn+11−xs_n =\displaystyle\frac{1-x^{n+1}}{1-x}

if x≠1x\neq1.

If 0<x<10< x<1 then xn+1→0x^{n+1}\to 0 as n→∞n\to\infty and we obtain sn→11−xs_n\to\displaystyle\frac{1}{1-x} so

∑n=0∞xn=11−x\displaystyle\sum_{n=0}^{\infty}x^n=\displaystyle\frac{1}{1-x}

The Exponential Function and the Poisson Distribution​

The exponential function can be written as a series (infinite sum):

ex=∑n=0∞xnn!e^x=\displaystyle\sum_{n=0}^{\infty}\displaystyle\frac{x^n}{n!}

The Poisson distribution is defined by the probabilities

p(x)=e−λλxx! for x=0, 1, 2, …p(x)=e^{-\lambda}\displaystyle\frac{\lambda^x}{x!}\textrm{ for } x=0,\ 1,\ 2,\ \ldots

Details​

The exponential function can be written as a series (infinite sum):

ex=∑n=0∞xnn!e^x=\displaystyle\sum_{n=0}^{\infty}\displaystyle\frac{x^n}{n!}

Knowing this we can see why the Poisson probabilities

p(x)=e−λλxx!p(x)=e^{-\lambda}\displaystyle\frac{\lambda^x}{x!}

add to one:

∑x=0∞p(x)=∑x=0∞e−λλxx!=e−λ∑x=0∞λxx!=e−λeλ=1\displaystyle\sum_{x=0}^{\infty}p(x)=\displaystyle\sum_{x=0}^{\infty}e^{-\lambda}\displaystyle\frac{\lambda^x}{x!}=e^{-\lambda}\displaystyle\sum_{x=0}^{\infty}\displaystyle\frac{\lambda^x}{x!}=e^{-\lambda}e^{\lambda}=1

Relation to Expected Values​

The expected value for the Poisson is given by

∑x=0∞xp(x)=∑x=0∞xe−λλxx!=λ\begin{aligned} \displaystyle\sum_{x=0}^\infty x p(x) &= \displaystyle\sum_{x=0}^\infty x e^{-\lambda} \displaystyle\frac{\lambda^x}{x!} \\ &= \lambda \end{aligned}

Details​

The expected value for the Poisson is given by

∑x=0∞xp(x)=∑x=0∞xe−λλxx!=e−λ∑x=1∞xλxx!=e−λ∑x=1∞λx(x−1)!=e−λλ∑x=1∞λ(x−1)(x−1)!=e−λλ∑x=0∞λxx!=e−λλeλ=λ\begin{aligned} \displaystyle\sum_{x=0}^\infty x p(x) &= \displaystyle\sum_{x=0}^\infty x e^{-\lambda} \displaystyle\frac{\lambda^x}{x!} \\ &= e^{-\lambda} \displaystyle\sum_{x=1}^\infty \displaystyle\frac{x\lambda^x}{x!} \\ &= e^{-\lambda} \displaystyle\sum_{x=1}^\infty \displaystyle\frac{\lambda^x}{(x-1)!} \\ &= e^{-\lambda} \lambda \displaystyle\sum_{x=1}^\infty \displaystyle\frac{\lambda^{(x-1)}}{(x-1)!} \\ &= e^{-\lambda} \lambda \displaystyle\sum_{x=0}^\infty \displaystyle\frac{\lambda^{x}}{x!} \\ &= e^{-\lambda} \lambda e^{\lambda} \\ &= \lambda \end{aligned}