Convergence of P-Series/Real

Theorem

Let $p \in \R$ be a real number.

Then the $p$-series:

$\displaystyle \sum_{n \mathop = 1}^\infty n^{-p}$

is convergent if and only if $p > 1$.

Proof 1

By the Integral Test:

$\displaystyle \sum_{n \mathop = 1}^\infty \frac 1 {n^x}$ converges if and only if the improper integral $\displaystyle \int_1^\infty \frac {\d t} {t^x}$ exists.

The result follows from Integral to Infinity of Reciprocal of Power of x.

$\blacksquare$

Proof 2

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:

 $\displaystyle s_N$ $\le$ $\displaystyle S_{2^N - 1}$ $\displaystyle$ $=$ $\displaystyle 1 + \dfrac 1 {2^p} + \dfrac 1 {3^p} + \dfrac 1 {4^p} + \dotsb + \dfrac 1 {\paren {2^N - 1}^p}$ $\displaystyle$ $=$ $\displaystyle 1 + \paren {\dfrac 1 {2^p} + \dfrac 1 {3^p} } + \paren {\dfrac 1 {4^p} + \dfrac 1 {5^p} + \dfrac 1 {6^p} + \dfrac 1 {7^p} } + \dotsb + \paren {\dfrac 1 {2^{\paren {N - 1} p} } + \dotsb + \dfrac 1 {\paren {2^N - 1}^p} }$ $\displaystyle$ $\le$ $\displaystyle 1 + \paren {\dfrac 1 {2^p} + \dfrac 1 {2^p} } + \paren {\dfrac 1 {4^p} + \dfrac 1 {4^p} + \dfrac 1 {4^p} + \dfrac 1 {4^p} } + \dotsb + \paren {\dfrac 1 {2^{\paren {N - 1} p} } + \dotsb + \dfrac 1 {2^{\paren {N - 1} p} } }$ $\displaystyle$ $=$ $\displaystyle 1 + \dfrac 2 {2^p} + \dfrac 4 {4^p} + \dotsb + \dfrac {2^{N - 1} } {2^{\paren {N - 1} p} }$ $\displaystyle$ $=$ $\displaystyle 1 + \dfrac 1 {2^{p - 1} } + \paren {\dfrac 1 {2^{p - 1} } }^2 + \dotsb + \paren {\dfrac 1 {2^{p - 1} } }^{n - 1}$ $\displaystyle$ $=$ $\displaystyle \dfrac {1 - \paren {1 / 2^{p - 1} }^N} {1 - \paren {1 / 2^{p - 1} } }$ $\displaystyle$ $\le$ $\displaystyle \dfrac 1 {1 - \paren {1 / 2^{p - 1} } }$

Hence the result.

$\blacksquare$