Convergence of P-Series/Lemma
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Lemma for Convergence of P-Series
Let $p = x + i y$ be a complex number where $x, y \in \R$ such that:
- $x > 0$
- $x \ne 1$
Then:
- $\ds \sum_{n \mathop = 1}^\infty \frac 1 {n^x}$ converges if and only if $\ds \lim_{P \mathop \to \infty} \dfrac {P^{1 - x} } {1 - x}$ converges.
Proof
Let $p = x + i y$.
Then:
\(\ds \sum_{n \mathop = 1}^\infty \size {n^{-p} }\) | \(=\) | \(\ds \sum_{n \mathop = 1}^\infty \frac 1 {\size {n^x n^{i y} } }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{n \mathop = 1}^\infty \frac 1 {\size {n^x e^{-i y \, \map \log n} } }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{n \mathop = 1}^\infty \frac 1 {\size {n^x} \size {e^{-i y \, \map \log n} } }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{n \mathop = 1}^\infty \frac 1 {\size {n^x} }\) |
by Euler's Formula.
Now since $x > 0$, and all $n \ge 1$, all terms are positive and we may do away with the absolute values.
Then by the Cauchy Integral Test:
- $\ds \sum_{n \mathop = 1}^{\to \infty} \frac 1 {n^x}$ converges if and only if $\ds \int_1^\infty \frac {\d t} {t^x}$ converges.
First let $x \ne 1$:
\(\ds \int_1^{\to \infty} \frac {\d t} {t^x}\) | \(=\) | \(\ds \lim_{P \mathop \to \infty} \int_1^P \dfrac {\d t} {t^x}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \lim_{P \mathop \to \infty} \intlimits {\dfrac {t^{1 - x} } {1 - x} } 1 P\) | Primitive of Power | |||||||||||
\(\ds \) | \(=\) | \(\ds \lim_{P \mathop \to \infty} \paren {\dfrac {P^{1 - x} } {1 - x} - \dfrac {1^{1 - x} } {1 - x} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \lim_{P \mathop \to \infty} \paren {\dfrac {P^{1 - x} } {1 - x} } - \dfrac {1^{1 - x} } {1 - x}\) |
Hence:
- $\ds \int_1^{\to \infty} \frac {\d t} {t^x} = \lim_{P \mathop \to \infty} \paren {\dfrac {P^{1 - x} } {1 - x} } - \dfrac {1^{1 - x} } {1 - x}$
- $\ds \int_1^{\to \infty} \frac {\d t} {t^x} + \dfrac {1^{1 - x} } {1 - x} = \lim_{P \mathop \to \infty} \paren {\dfrac {P^{1 - x} } {1 - x} }$
and so:
- $\ds \sum_{n \mathop = 1}^\infty \size {n^{-p} }$ converges
- $\ds \lim_{P \mathop \to \infty} \dfrac {P^{1 - x} } {1 - x}$ converges.
For $x = 1$:
\(\ds \int_1^{\to \infty} \frac {\d t} t\) | \(=\) | \(\ds \lim_{P \mathop \to \infty} \int_1^P \dfrac {\d t} t\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \lim_{P \mathop \to \infty} \bigintlimits {\ln x} 1 P\) | Primitive of Reciprocal | |||||||||||
\(\ds \) | \(=\) | \(\ds \lim_{P \mathop \to \infty} \ln P\) | Logarithm of 1 is 0 |
which diverges.
$\blacksquare$
Sources
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- 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 6.6$
- 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 13.34 \ (3)$
- 1992: Larry C. Andrews: Special Functions of Mathematics for Engineers (2nd ed.) ... (previous) ... (next): $\S 1.2.2$: Summary of convergence tests