Question

Prove using integral test that the Riemann zeta function converges when p > 1 and diverges whenever 0 < p ≤ 1.

Reimann zeta function:-
$\boxed{ \zeta(p) = \sum \limits_{n = 1}^ \infty \dfrac{1}{ {n}^{ p} } = \frac{1}{ {1}^{p} } + \frac{1}{ {2}^{p} } + \frac{1}{ {3}^{p} } + ... }$
​

Answer 1

SOLUTION

TO DETERMINE

The below series converges when p > 1 and diverges whenever 0 < p ≤ 1

$\displaystyle \sf \sum\limits_{n=1}^{ \infty } \: \frac{1}{ {n}^{p} }$

PROOF

1st Method :

Using Integral Test

Integral test state that a positive term series f(1) + f(2) +.. .. + f(n) + .. .. where f(n) decreases as n increases , converges or diverges according as the integral

$\displaystyle \sf \: \int\limits_{1}^{\infty}f(x)\, dx \: \: is \: finite \: or \: infinite$

By the above test the given seris will converge or diverge according as

$\displaystyle \sf \: \int\limits_{1}^{\infty} \: \frac{1}{ {x}^{p} } \, dx \: \: is \: finite \: or \: infinite$

Case : I

$\sf \: If \: p \ne \: 1$

$\displaystyle \sf \: \int\limits_{1}^{\infty} \: \frac{1}{ {x}^{p} } \, dx \:$

$\displaystyle \sf = \lim_{m \to \infty } \: \int\limits_{1}^{m} \: \frac{1}{ {x}^{p} } \, dx \:$

$\displaystyle \sf = \lim_{m \to \infty } \: \bigg( \frac{ {m}^{1 - p} - 1}{1 - p} \bigg)$

$= \begin{cases} & \sf{ \: \: \: \: \dfrac{1}{p - 1} \: \: \: \: when \: p > 1} \\ \\ & \sf{ \to \: \infty \: \: \: \: \: when \: p < 1} \end{cases}$

Case : II

For p = 1

$\displaystyle \sf \: \int\limits_{1}^{\infty} \: \frac{1}{ {x}^{p} } \, dx$

$\displaystyle = \sf \: \int\limits_{1}^{\infty} \: \frac{1}{ {x}^{} } \, dx$

$\displaystyle \sf \: = \log x \bigg| _{1}^{\infty}$

$\displaystyle \sf \to \: \: \infty$

Hence the proof follows

2nd Method :

Case : I

Let us check for p > 1

$\displaystyle \sf Let \: a_n = \: \frac{1}{ {n}^{p} }$

$\sf{Let \: \sum b_n \: is \: obtained \: by \: grouping \: the \: terms \: of \: \sum a_n}$

$\displaystyle \sf \sum a_n = 1 + \bigg( \frac{1}{ {2}^{p} } + \frac{1}{ {3}^{p} } \bigg) + \bigg( \frac{1}{ {4}^{p} } + \frac{1}{ {5}^{p} } + \frac{1}{ {6}^{p} } + \frac{1}{ {7}^{p} } \bigg) + .. \: ..$

Let us take

$\displaystyle \sf b_1 = 1$

$\displaystyle \sf b_2= \bigg( \frac{1}{ {2}^{p} } + \frac{1}{ {3}^{p} } \bigg)$

$\displaystyle \sf b_3 = \bigg( \frac{1}{ {4}^{p} } + \frac{1}{ {5}^{p} } + \frac{1}{ {6}^{p} } + \frac{1}{ {7}^{p} } \bigg)$

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.

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So on

We now observe that

$\displaystyle \sf b_2 < \bigg( \frac{1}{ {2}^{p} } + \frac{1}{ {2}^{p} }\bigg) = \frac{2}{ {2}^{p} } = \frac{1}{ {2}^{p - 1} }$

$\displaystyle \sf b_3 < \bigg( \frac{1}{ {4}^{p} } + \frac{1}{ {4}^{p} } +\frac{1}{ {4}^{p} } + \frac{1}{ {4}^{p} } \bigg) = \frac{4}{ {4}^{p} } = \frac{1}{{( {2}^{p - 1}) }^{2} }$

.

.

So on

In general

$\displaystyle \sf b_n < { \bigg( \frac{1}{{2}^{p - 1} } \bigg)}^{n - 1} \: \: \: \: for \: all \: n \geqslant 2$

$\displaystyle \sf Now \: when \: \: p > 1 \: we \: have \: \frac{1}{ {2}^{p - 1} } < 1$

$\displaystyle \sf \because \: \sum \: { \bigg( \frac{1}{{2}^{p - 1} } \bigg)}^{n - 1} \: is \: convergent \: geometric \: series \: as \: common \: ratio \: < 1$

$\sf \therefore \: \sum b_n \: \: is \: convergent \: by \: comparision \: test$

$\sf \therefore \: \sum a_n \: \: is \: convergent \:$

$\displaystyle \sf \therefore \: \sum\limits_{n=1}^{ \infty } \: \frac{1}{ {n}^{p} } \: \: is \: convergent$

Case : II

Let us check for p = 1

Putting p = 1 we get the Harmonic series which is divergent by Cauchy's Principle

Case : III

Let us check for p < 1

Then we have

$\displaystyle \sf \frac{1}{ {2}^{p} } > \frac{1}{2}$

$\displaystyle \sf \frac{1}{ {3}^{p} } > \frac{1}{3}$

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So on

$\displaystyle \sf \because \: \sum\limits_{n=1}^{ \infty } \: \frac{1}{ {n}^{} } \: \: is \: divergent \: \: Harmonic \: series$

$\sf \therefore \: \sum a_n \: \: is \: divergent \:$

$\displaystyle \sf \therefore \: \sum\limits_{n=1}^{ \infty } \: \frac{1}{ {n}^{p} } \: \: is \: convergent$

Hence the proof follows

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