Convergence of P-Series/Real/Proof 2
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Theorem
Let $p \in \R$ be a real number.
Then the $p$-series:
- $\ds \sum_{n \mathop = 1}^\infty n^{-p}$
is convergent if and only if $p > 1$.
Proof
Let $p = 1$.
Then from Harmonic Series is Divergent the $p$-series diverges.
So let $p > 1$.
We note that the sequence of partial sums is increasing.
Hence it is sufficient to show that they are bounded above.
Let:
- $s_{2^N} := 1 + \dfrac 1 {2^p} + \dfrac 1 {3^p} + \dotsb + \dfrac 1 {N^p}$
Then:
\(\ds s_N\) | \(\le\) | \(\ds S_{2^N - 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 1 + \dfrac 1 {2^p} + \dfrac 1 {3^p} + \dfrac 1 {4^p} + \dotsb + \dfrac 1 {\paren {2^N - 1}^p}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 1 + \paren {\dfrac 1 {2^p} + \dfrac 1 {3^p} } + \paren {\dfrac 1 {4^p} + \dfrac 1 {5^p} + \dfrac 1 {6^p} + \dfrac 1 {7^p} } + \dotsb + \paren {\dfrac 1 {2^{\paren {N - 1} p} } + \dotsb + \dfrac 1 {\paren {2^N - 1}^p} }\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds 1 + \paren {\dfrac 1 {2^p} + \dfrac 1 {2^p} } + \paren {\dfrac 1 {4^p} + \dfrac 1 {4^p} + \dfrac 1 {4^p} + \dfrac 1 {4^p} } + \dotsb + \paren {\dfrac 1 {2^{\paren {N - 1} p} } + \dotsb + \dfrac 1 {2^{\paren {N - 1} p} } }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 1 + \dfrac 2 {2^p} + \dfrac 4 {4^p} + \dotsb + \dfrac {2^{N - 1} } {2^{\paren {N - 1} p} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 1 + \dfrac 1 {2^{p - 1} } + \paren {\dfrac 1 {2^{p - 1} } }^2 + \dotsb + \paren {\dfrac 1 {2^{p - 1} } }^{n - 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {1 - \paren {1 / 2^{p - 1} }^N} {1 - \paren {1 / 2^{p - 1} } }\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds \dfrac 1 {1 - \paren {1 / 2^{p - 1} } }\) |
Hence the result.
$\blacksquare$
Sources
- 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 6.6$