Power Series Expansion for Tangent Function/Proof of Convergence
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Theorem
The radius of convergence of the Power Series Expansion for Tangent Function:
- $\ds \tan x = \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n - 1} 2^{2 n} \paren {2^{2 n} - 1} B_{2 n} } {\paren {2 n}!} x^{2 n - 1}$
where $B_{2 n}$ denotes the Bernoulli numbers, is given as:
- $\size x < \dfrac \pi 2$
Proof
By Combination Theorem for Limits of Real Functions we can deduce the following.
| \(\ds \) | \(\) | \(\ds \lim_{n \mathop \to \infty} \size {\frac {\frac {\paren {-1}^n 2^{2 n + 2} \paren {2^{2 n + 2} - 1} B_{2 n + 2} } {\paren {2 n + 2}!} x^{2 n + 1} } {\frac {\paren {-1}^{n - 1} 2^{2 n} \paren {2^{2 n} - 1} B_{2 n} } {\paren {2 n}!} x^{2 n - 1} } }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \lim_{n \mathop \to \infty} \size {\frac {\paren {2^{2 n + 2} - 1} } {\paren {2^{2 n} - 1} } \frac 1 {\paren {2 n + 1} \paren {2 n + 2} } \frac {B_{2 n + 2} } {B_{2 n} } } 4 x^2\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \lim_{n \mathop \to \infty} \size {\frac {2^{2 n + 2} - 1} {2^{2 n} - 1} } \size {\frac 1 {\paren {2 n + 1} \paren {n + 1} } \frac {B_{2 n + 2} } {B_{2 n} } } 2 x^2\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \lim_{n \mathop \to \infty} \size {4 \frac {2^{2 n} } {2^{2 n} - 1} - \frac 1 {2^{2 n} - 1} } \size {\frac 1 {\paren {2 n + 1} \paren {n + 1} } \frac {B_{2 n + 2} } {B_{2 n} } } 2 x^2\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \lim_{n \mathop \to \infty} \size {4 + \frac 4 {2^{2 n} - 1} - \frac 1 {2^{2 n} - 1} } \size {\frac 1 {\paren {2 n + 1} \paren {n + 1} } \frac {B_{2 n + 2} } {B_{2 n} } } 2 x^2\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \lim_{n \mathop \to \infty} \size {\frac 1 {\paren {2 n + 1} \paren {n + 1} } \frac {B_{2 n + 2} } { B_{2 n} } } 8 x^2\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \lim_{n \mathop \to \infty} \size {\frac 1 {\paren {2 n + 1} \paren {n + 1} } \frac {\paren {-1}^{n + 2} 4 \sqrt {\pi \paren {n + 1} } \paren {\frac {n + 1} {\pi e} }^{2 n + 2} } {\paren {-1}^{n + 1} 4 \sqrt {\pi n} \paren {\frac n {\pi e} }^{2 n} } } 8 x^2\) | Asymptotic Formula for Bernoulli Numbers | |||||||||||
| \(\ds \) | \(=\) | \(\ds \lim_{n \mathop \to \infty} \size {\frac {\paren {n + 1}^2} {\paren {2 n + 1} \paren {n + 1} } \sqrt {\frac {n + 1} n } \paren {\frac {n + 1} n}^{2 n} } \frac 8 {\pi^2 e^2} x^2\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \lim_{n \mathop \to \infty} \size {\paren {\frac {n + 1} n}^{2 n} } \frac 4 {\pi^2 e^2} x^2\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \lim_{n \mathop \to \infty} \size {\paren {\paren {1 + \frac 1 n}^n}^2} \frac 4 {\pi^2 e^2} x^2\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {4 e^2} {\pi^2 e^2} x^2\) | Definition of Euler's Number as Limit of Sequence | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac 4 {\pi^2} x^2\) |
This is less than $1$ if and only if:
- $\size x < \dfrac \pi 2$
Hence by the Ratio Test, the series converges for $\size x < \dfrac \pi 2$.
$\blacksquare$