Convergence of P-Series/Lemma

From ProofWiki
Jump to navigation Jump to search

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:

$\ds \int_1^{\to \infty} \frac {\d t} {t^x} = \lim_{P \mathop \to \infty} \paren {\dfrac {P^{1 - x} } {1 - x} } - \dfrac {1^{1 - x} } {1 - x}$

if and only if:

$\ds \int_1^{\to \infty} \frac {\d t} {t^x} + \dfrac {1^{1 - x} } {1 - x} = \lim_{P \mathop \to \infty} \paren {\dfrac {P^{1 - x} } {1 - x} }$


and so:

$\ds \sum_{n \mathop = 1}^\infty \size {n^{-p} }$ converges

if and only if:

$\ds \lim_{P \mathop \to \infty} \dfrac {P^{1 - x} } {1 - x}$ 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


This page may be the result of a refactoring operation.
As such, the following source works, along with any process flow, will need to be reviewed.
When this has been completed, the citation of that source work (if it is appropriate that it stay on this page) is to be placed above this message, into the usual chronological ordering.
If you have access to any of these works, then you are invited to review this list, and make any necessary corrections.
To discuss this page in more detail, feel free to use the talk page.
When this work has been completed, you may remove this instance of {{SourceReview}} from the code.


Retrieved from "https://proofwiki.org/w/index.php?title=Convergence_of_P-Series/Lemma&oldid=654781"