Question

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

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

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Answer 1

Hint: In order to apply integral test on an infinite decreasing series, find a function such that $f(n) = a_n$ and find it's Integral from 1 to ∞, if the integral results a finite value, then only the sum converges to a finite value.

Solution:-

Firstly we have to define a function such that $f(n) = a_n$.

$\longrightarrow f(n) = a_n \implies f(n) = \dfrac{1}{n^{p} }$

$\boxed{\longrightarrow f(x) = \frac{1}{x^p} }$

Now finding the integral of this function from 1 to infinity.

${\longrightarrow \int_1^\infty f(x) dx \implies \int_1^\infty \dfrac{1}{x^p}dx \implies \boxed{\int_1^\infty x^{-p} dx}}$

Now, we have three cases- when p=1, when 0 < p < 1 and last one when p > 1.

If p = 1

$\displaystyle I = \int_1^\infty x^{-p} dx\\\\\implies I = \int_1^\infty x^{-1} dx\\\\\implies I = \left[x^{-1}\right]^\infty_1 \\\\\implies I = \ln(\infty) - \ln(1)\\\\\implies I = \infty$

Since the integral results an infinite value, the given zeta function will diverge to infinity.

If 0 < p < 1

$\displaystyle I = \int_1^\infty x^{-p} dx\\\\\implies I = \left[\frac{x^{-p+1}}{-p+1}\right]^\infty_1\\\\\implies I = \dfrac{\infty}{\mathbb{R}^+}\\\\\implies I = \infty$

Again the integral results an infinite value, therefore the zeta function diverges to infinity.

If p > 1

$\displaystyle I = \int_1^\infty x^{-p} dx\\\\\implies I = \left[\dfrac{x^{-p+1}}{-p+1}\right]^\infty_1\\\\\implies I = \left[\frac{\frac{x}{x^p}}{1-p}\right]^\infty_1\\\\\implies I = \left[\frac{\frac{1}{x^{p-1}}}{1-p}\right]^\infty_1\\\\\implies I = \left[\dfrac{1}{(x^{p-1})(1-p)}\right]^\infty_1\\\\\implies I = \left[\dfrac{1}{\infty\times \mathbb{R}^-} - \dfrac{1}{\mathbb{R^-}}\right]\\\\\implies I = 0 - (-\mathbb R^+)\\\\\implies I = \mathbb{R}^+$

Therefore, the given zeta function will converge to a finite real value as p > 1.

We have proved the required result.

*Refer to the graphs for verification, three graphs have been attached where 0 < p < 1, p > 1 and p = 1. Feel free to contact if you have any problem :)

Answer 2

To prove using the integral test that the Riemann zeta function converges when p > 1 and diverges whenever 0 < p ≤ 1, we need to compare the series with an appropriate integral.

Let's consider the function $\rm f(x) = x^{-p}$. This function is positive, continuous, and decreasing for $\rm x \geq 1$. Therefore, we can use the integral test to compare the series $\rm\zeta(p) =\rm \sum\limits_{n=1}^\infty \frac{1}{n^p}$ with the integral $\rm\int_1^\infty f(x)dx = \int_1^\infty x^{-p}dx$.

If the integral converges, then the series also converges. If the integral diverges, then the series also diverges.

Case 1: p > 1

In this case, we have:

$\rm\int_1^\infty x^{-p}dx = \lim\limits_{t\to\infty} \left[\frac{x^{1-p}}{1-p}\right]_1^t = \frac{1}{p-1} < \infty$

Since the integral converges, the series also converges by the integral test. Therefore, the Riemann zeta function converges when p > 1.

Case 2: 0 < p ≤ 1

In this case, we have:

$\rm\int_1^\infty x^{-p}dx = \lim\limits_{t\to\infty} \left[\frac{x^{1-p}}{1-p}\right]_1^t = \lim\limits_{t\to\infty} \frac{t^{1-p} - 1}{1-p}$

If p = 1, the limit is infinity. If 0 < p < 1, the limit is zero. Therefore, the integral diverges if p = 1 and converges if 0 < p < 1.

Since the integral diverges whenever 0 < p ≤ 1, the series also diverges by the integral test. Therefore, the Riemann zeta function diverges whenever 0 < p ≤ 1.

Therefore, we have proven using the integral test that the Riemann zeta function converges when p > 1 and diverges whenever 0 < p ≤ 1.