Power Series
Definition
Let (a_n) be a sequence of real numbers. The series
\sum^\infty_{n=0} a_n x^n = a_0 + a_1x + a_2x^2 + \cdots
Is called the power series of a_n.
Power series can be thought of as a generalization of polynomials, because given any polynomial
P = a_0 + a_1 x + a_2 x^2 + \cdots + a_n x^n
We can define a sequence
(a_n) = (a_0,a_1,a_2,\cdots,a_n,0,0,0,\cdots)
And then the power series of (a_n) will be equal to the polynomial P.
We let \mathbb{R}[x] denote a polynomial with real coeffients, and \mathbb{R}[|x|] denote a power series with real coeffients.
Let (a_n) be a sequence of real numbers. Let \alpha = \limsup |a_n|^{1/n}. We define a function R such that
R = \begin{cases} \frac{1}{\alpha} & \text{if } 0 < \alpha < \infty\\ \infty & \text{if } \alpha = 0\\ 0& \text{if } \alpha = \infty \end{cases}
The power series of (a_n)
(a) Converges absolutly whenever |x| < R, and
(b) Diverges when |x| > R.
Let \sum a_n and \sum b_n be infinite series such that a_n,b_n \leq 0 for all n. Then
(a) If \sum a_n converges and 0 \leq b_n \leq a_n for all n, then \sum b_n converges.
(b) If \sum a_n = +and 0 \leq a_n \leq b_n for all n, then \sum b_n = + \infty.
(a) Because s_{a_n} is monotone and bounded, s_{b_n} converges.
If \sum |a_n| converges, then \sum a_n is absolutly convergent, or that the sum converges absolutly.
If a series converges absolutly, then it converges.
If \sum a_n, a_n \not = 0
(a) If \limsup |a_{n+1}/a_n| < 1, the the series is absolutly convergent.
(b) If \liminf |a_{n+1}/a_n| > 1 then the series diverges.
Let f be an integerable function defined on [0,\infty) and suppose that f is positive and decreasing.
Then the series \sum f(n) converges if and only if
\lim_{n \to \infty} \left( \int^n_1 f(x) \,dx \right)
exists, and is finite.
Direct Proof:
Let a_n = f(n), and b_n = \int^{n+1}_n f(x)\,dx.
Since f is decreasing, given an integer n, f(n+1) \leq f(x) \leq f(n) for all x \in [n,n+1].
By Theorem 444, we have that
\left| \int^{n+1}_{n} \right| \leq \int^{n+1}_{n} |f| \leq |f|.
Therefore we have 0 \leq a_{n+1} \leq b_n \leq a_n for all integers n.
By applying the comparison test twice, we have that \sum a_n converges if and only if \sum b_n converges. Note that
\sum^k_{i=1} b_i = \int^{k+1}_1 f
Given the series
\sum^{\infty}_{n=0} ne^{-n^2},
we define a function f(x) = x\exp(-x^2). If f is integerable, positive, and decreasing on [k,\infty), then we can use the integral test.
The function is continuous, and positive by inspection. Taking the derivative of the function gets us
f^\prime(x) = e^{-x^2}(1-2x^2).
Since f^\prime is non-positive on [1/\sqrt{2},\infty), we can apply the integral test. We calculate
\lim_{t\to\infty}\left( \int^t_1 f \right) = \frac{1}{2}
If (a_n) is a decreasing sequence of positive numbers and \lim a_n = 0, then the series \sum (-1)^{n}a_n converges.
Let (f_n) be a sequence of functions defined on S \subseteq \mathbb{R}. The sequence is pointwise convergent if for each x \in S, we have that (f_n(x)) converges.
Let (f_n) be a pointwise convergent sequence. Define f \colon S \to \mathbb{R} such that f(x) = \lim f_n(x). If f is continuous, then (f_n) is uniformly convergent
Let (f_n) be sequence of functions defined on S \subseteq \mathbb{R}. (f_n) converges uniformly to f on s if
\forall \varepsilon > 0, \exists N \in \mathbb{N} s.t. \forall n \geq N and \forall x \in S, |f_n(x) - f(x)| < \varepsilon.
Let (f_n) be a sequence of functions on S \subseteq \mathbb{R} such tha
Given \varepsilon > 0, by the assumption we know that \exists N \in \mathbb{N} such that |f_n(x) - f_m(x)| < \varepsilon/2 for all x \in S and all m,n\geq N.
Choose some n \geq N. Since f_m \to \infty.
Let (f_n) be a sequence of functions. The series \sum f_n is said to converge pointwise if the sequence of partial sums
s_n \coloneq \sum^n_{k=0}f_k(x)
Let (f_n) be a sequence of functions. Let (M_n) be a sequence of real numbers such that |f_n(x)| \leq M_n for all x\in S and for all n. If \sum M_n converges, then \sum f_n converges uniformly on S.
Since \sum M_n converges, we have that \exists N \in \mathbb{N} such that m,n \geq N implies |s_n - s_{m-1}| < \varepsilon. If n \geq m \geq N, then
|s_n(x) - s_m(x)|
Given \varepsilon = 1, let N \in \mathbb{N} we need to find som