Theorem 24. Let \(\sum_{n=0}^\infty a_n (x-x_0)^n\) be a power series.
We set
\(R := \frac{1}{\limsup_{n \to \infty} \sqrt[n]{|a_n|}}\),
where we interpret \(\frac{1}{\infty}\) as 0, and \(\frac{1}{0}\) as \(\infty\)
(i.e. \(R \in [0, \infty) \cup \{\infty\}\)).
If \(R = \infty\), then the power series is absolutely convergent for every \(x \in \mathbb{R}\).
If \(R \in (0, \infty)\), then the power series is
\(\begin{cases} \text{absolutely convergent} & \text{if } |x-x_0| < R \\ \text{divergent} & \text{if } |x-x_0|> R \end{cases}\).
Proof. We apply the root test. If \(R = \infty\),
i.e. \(\limsup_{n \to \infty} \sqrt[n]{|a_n|} = 0\), we have
\(\limsup \sqrt[n]{|a_n| |x-x_0|^n} = \limsup \sqrt[n]{|a_n|} \cdot |x-x_0| = 0 < 1\),
so the power series is abs. convergent for every \(x \in \mathbb{R}\).
If \(R \in (0, \infty)\), i.e. \(\limsup_{n \to \infty} \sqrt[n]{|a_n|} \in (0, \infty)\),
then we have
\(\limsup \sqrt[n]{|a_n| |x-x_0|^n} = \frac{1}{R} |x-x_0|\)
\(\begin{cases} < 1 \text{ if } |x-x_0| < R \implies \text{ abs. convergence} \\> 1 \text{ if } |x-x_0| > R \implies \text{ divergence} \end{cases}\)
Definition. The number \(R\) in Thm 24 is called the radius of convergence of the power series \(\sum_{n=0}^\infty a_n (x-x_0)^n\).
Remarks. i) In the case \(R = 0\) the power series only converges at \(x = x_0\). For example, the series \(\sum_{n=0}^\infty n^n x^n\) has radius of convergence 0.
ii) In the case \(R \in (0, \infty)\) the power series is absolutely convergent inside the "disc" \(U_R(x_0)\)
and divergent for all \(x \in (-\infty, x_0-R) \cup (x_0+R, \infty)\). For the boundary points with \(|x-x_0| = R\) there is no general statement on convergence.
iii) An alternative way to compute the radius of convergence of a power series \(\sum_{n=0}^\infty a_n (x-x_0)^n\) is using the ratio test, if possible:
If \(a_n \neq 0\) for \(n \ge N\) (\(N \in \mathbb{N}\) fixed) and if \(\lim_{n \to \infty} |\frac{a_n}{a_{n+1}}|\) exists or is \(\infty\), then for the radius of convergence \(R\) we have
\(R = \frac{1}{\limsup_{n \to \infty} \sqrt[n]{|a_n|}} = \lim_{n \to \infty} |\frac{a_n}{a_{n+1}}|\).
It follows, by the way, that for every sequence \((a_n)_{n=0}^\infty\) such that \(\lim_{n \to \infty} |\frac{a_{n+1}}{a_n}|\) exists or is \(\infty\), we have
\(\limsup_{n \to \infty} \sqrt[n]{|a_n|} = \lim_{n \to \infty} |\frac{a_{n+1}}{a_n}|\).
Examples. i) For \(\sum_{n=0}^\infty x^n\), i.e. \(a_n = 1, n \ge 0\), we have \(R = \frac{1}{\limsup \sqrt[n]{1}} = 1\).
ii) For \(\sum_{n=0}^\infty \frac{x^n}{n!}\), i.e. \(a_n = \frac{1}{n!}, n \ge 0\), we have
\(\lim_{n \to \infty} |\frac{a_n}{a_{n+1}}| = \lim_{n \to \infty} \frac{1/n!}{1/(n+1)!} = \lim_{n \to \infty} (n+1) = \infty\),
so the series \(\sum_{n=0}^\infty \frac{x^n}{n!}\) is absolutely convergent for every \(x \in \mathbb{R}\).
iii) For \(\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} x^n\), i.e. \(a_n = \frac{(-1)^{n+1}}{n}\) for \(n \ge 1\) and \(a_0 = 0\), we have
\(\limsup_{n \to \infty} \sqrt[n]{\frac{1}{n}} = \lim_{n \to \infty} \frac{1}{\sqrt[n]{n}} = 1\), so
the series converges absolutely for \(|x| < 1\).
iv) The power series \(\sum_{n=0}^\infty \frac{(-1)^n}{(2n)!} x^{2n}\) and \(\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!} x^{2n+1}\) both are absolutely convergent for all \(x \in \mathbb{R}\).
Next we study important functions.
Theorem 25. For every \(x \in \mathbb{R}\) we have
\(\lim_{n \to \infty} (1 + \frac{x}{n})^n = \sum_{k=0}^\infty \frac{x^k}{k!}\).
Proof. As \(\lim_{n \to \infty} \sum_{k=0}^n \frac{x^k}{k!} = \sum_{k=0}^\infty \frac{x^k}{k!}\), it is sufficient to show \(\lim_{n \to \infty} \{ (1 + \frac{x}{n})^n - \sum_{k=0}^n \frac{x^k}{k!} \} = 0\).
For \(n > 0\) we have
\(| \sum_{k=0}^n \frac{x^k}{k!} - (1 + \frac{x}{n})^n | \stackrel{\text{Binom.}}{=} | \sum_{k=0}^n (\frac{x^k}{k!} - \binom{n}{k} \frac{x^k}{n^k}) |\)
\(\le \sum_{k=2}^n \frac{|x|^k}{k!} (1 - \frac{n(n-1) \dots (n-k+1)}{n^k})\)
\(\le 1 - \frac{(n-k+1)^k}{n^k} = 1 - (1 - \frac{k-1}{n})^k\)
\(\stackrel{\text{Bernoulli}}{\le} 1 - (1 - \frac{k(k-1)}{n}) = \frac{k(k-1)}{n}\)
\(\le \sum_{k=2}^n \frac{|x|^k}{k!} \frac{k(k-1)}{n} = \frac{1}{n} \sum_{k=2}^n \frac{|x|^k}{(k-2)!}\)
\(\le \frac{|x|^2}{n} \sum_{k=0}^n \frac{|x|^k}{k!} \to 0, n \to \infty\).
Definition. The exponential function \(\exp: \mathbb{R} \to \mathbb{R}\) is defined by
\(\exp(x) := \sum_{k=0}^\infty \frac{x^k}{k!}\).
The series is called exponential series. Instead of \(\exp(x)\) we also write \(e^x\).
Theorem 26. For all \(x, y \in \mathbb{R}\) we have
i) \(e^{x+y} = e^x e^y\) "functional equation"
ii) \(e^x > 0, e^{-x} = \frac{1}{e^x}\),
iii) \(e^x \ge 1+x\), \(e^x \le \frac{1}{1-x}\) if \(x < 1\),
iv) \(\exp\) is strictly increasing on \(\mathbb{R}\), i.e.,
\(e^x < e^y\) if \(x < y\),
v) If \((x_n)_{n=0}^\infty\) is a sequence with \(\lim_{n \to \infty} x_n = x\),
then \(\lim_{n \to \infty} e^{x_n} = e^x\).
vi) \(\exp: \mathbb{R} \to (0, \infty)\) is bijective.
Proof. i) We have for \(x, y \in \mathbb{R}\)
\(e^x e^y = (\sum_{k=0}^\infty \frac{x^k}{k!}) (\sum_{j=0}^\infty \frac{y^j}{j!}) \stackrel{\text{Thm 22}}{=} \sum_{k=0}^\infty (\sum_{j=0}^k \frac{x^j}{j!} \frac{y^{k-j}}{(k-j)!})\)
\(= \sum_{k=0}^\infty \frac{1}{k!} (\sum_{j=0}^k \binom{k}{j} x^j y^{k-j}) = \sum_{k=0}^\infty \frac{(x+y)^k}{k!} = e^{x+y}\).
ii) We have for \(x \in \mathbb{R}\): \(1 = e^0 \stackrel{i)}{=} e^{x-x} = e^x e^{-x}\)
\(\implies e^{-x} = \frac{1}{e^x}\).
Moreover, for \(x \ge 0\): \(e^x = \sum_{k=0}^\infty \frac{x^k}{k!} \ge 1 > 0\)
and for \(x < 0\): \(e^x=\frac{1}{e^{-x}}> 0\).
iii) We know for \(x \ge -1\) and \(n \in \mathbb{N}\)
\((1 + \frac{x}{n})^n \stackrel{\text{Bernoulli}}{\ge} 1+x \implies e^x = \lim_{n \to \infty} (1 + \frac{x}{n})^n \ge 1+x\),
and for \(x < -1\): \(e^x> 0 > 1+x\).
Moreover, for \(x < 1\), we have
\(e^{-x} \ge 1-x > 0 \implies e^x \le \frac{1}{1-x}\).
iv) For \(x < y\) we have
\(e^{y-x} \stackrel{iii)}{\ge} 1+y-x > 1 \implies e^y > e^x\).
v) If \(x_n \to x\), then we know
\(\exists N \in \mathbb{N} \forall n \ge N: |x_n - x| < 1\).
\(\implies \forall n \ge N: e^{x_n} = e^{x_n - x + x} \stackrel{i)}{=} e^{x_n - x} e^x\)
\(\stackrel{iii)}{\le} e^x \frac{1}{1-(x_n-x)} \to e^x, n \to \infty\),
and \(e^{x_n} = e^x e^{x_n-x} \stackrel{iii)}{\ge} e^x (1+x_n-x) \to e^x, n \to \infty\).
\(\implies\) (Squeeze theorem) \(e^{x_n} \to e^x, n \to \infty\).
vi) The injectivity follows from part iv), so we only need to prove the surjectivity, i.e. \(\forall y \in (0, \infty) \exists x \in \mathbb{R}: e^x = y\).
Let \(y \in (0, \infty)\). We set \(S := \{ z \in \mathbb{R} : e^z \le y \}\).
It follows from iii) that \(\lim_{n \to \infty} e^{-n} = 0\), so \(S \neq \emptyset\). Moreover, for every \(z \in S\)
\(y \ge e^z \ge 1+z \implies z \le y-1\),
so \(S\) is bounded above. Hence, \(\sup S \in \mathbb{R}\). We define \(x := \sup S\), and we show \(e^x = y\).
To this end, we observe that
\(\exists x_n \to x\) with \(x_n \in S\) for all \(n \ge 1\).
\(\stackrel{v), \text{Thm 4}}{\implies} e^x = \lim_{n \to \infty} e^{x_n} \le y\).
Assume: \(e^x < y \implies 1 < e^{-x} y\)
We know from iv) and v) that \(1 < e^{1/n}\) for all \(n \ge 1\) and \(e^{1/n} \to e^0=1, n \to \infty\)
\(\implies \exists N \in \mathbb{N} \forall n \ge N : 1 < e^{1/n} < y e^{-x}\)
\(\implies \forall n \ge N : e^{x+1/n} < y\), which is a contradiction to \(x=\sup S\).
Hence, we have \(e^x = y\).
Definition. The inverse function of \(\exp: \mathbb{R} \to (0, \infty)\) is called (natural) logarithm and we denote it by \(\log: (0, \infty) \to \mathbb{R}\).
Remark. For \(x \in \mathbb{R}\) we have \(\log(e^x) = x\), and for \(x > 0\) we have \(e^{\log x} = x\).
Theorem 27. We have for \(x, y > 0\)
i) \(\log(xy) = \log x + \log y\) "functional equation"
ii) \(\log\) is strictly increasing on \((0, \infty)\), i.e.,
\(\log(x) < \log(y)\) if \(x < y\).
Proof. i) We show \(e^{\log(xy)} = e^{\log x + \log y}\):
\(e^{\log x + \log y} \stackrel{\text{Thm 26}}{=} e^{\log x} e^{\log y} = xy = e^{\log(xy)}\).
ii) Let \(0 < x < y \stackrel{\text{Thm 26}}{\implies} \exists a, b \in \mathbb{R}\) such that \(a < b\) and \(x=e^a, y=e^b\).
\(\implies \log x = \log(e^a) = a < b=\log(e^b)=\log y\).
Next we define general powers.
Let \(a > 0\) and \(n \in \mathbb{N}\), then
\(a^n = a \cdot a \dots a = e^{\log a} \cdot e^{\log a} \dots e^{\log a} \stackrel{\text{Thm 26}}{=} e^{n \log a}\).
Definition. For \(a > 0\) and \(x \in \mathbb{R}\) we define
\(a^x := e^{x \log a}\),
where \(a\) is the base and \(x\) is the power or the exponent.
Moreover, the logarithm with base a is defined by
\(\log_a x := \frac{\log x}{\log a}\), for \(a > 0, a \neq 1\) and \(x > 0\).
Remark. The natural logarithm is the logarithm with base \(e\), and as \(\log x\) is the inverse function of \(e^x\), \(\log_a x\) is the inverse function of \(a^x\).
Theorem 28. For \(a, b > 0\) and \(x, y \in \mathbb{R}\) we have
i) \(a^x a^y = a^{x+y}\)
ii) \((a^x)^y = a^{xy}\)
iii) \(a^x b^x = (ab)^x\)
iv) If \(a > 1\), then we have
\(x < y \implies a^x < a^y\),
and if \(a < 1\), then we have
\(x < y \implies a^x> a^y\).
Proof. Follows immediately from Thm 26, Thm 27 and the definition of general powers.