The series does not converge

Real Analysis
Post Reply
User avatar
Tolaso J Kos
Administrator
Administrator
Posts: 867
Joined: Sat Nov 07, 2015 6:12 pm
Location: Larisa
Contact:

The series does not converge

#1

Post by Tolaso J Kos »

Of course the following is obvious , but let us see a proof of this:

Prove that the series: \( \displaystyle \sum_{n=1}^{\infty}\frac{1}{n^k},\,\,\,\,k\in [0, 1] \) does not converge.

I have a (nice) proof for this one. Let me see some other points of view.
Imagination is much more important than knowledge.
Papapetros Vaggelis
Community Team
Posts: 426
Joined: Mon Nov 09, 2015 1:52 pm

Re: The series does not converge

#2

Post by Papapetros Vaggelis »

If \(\displaystyle{k=0}\) , then :

\(\displaystyle{\begin{aligned} \sum_{n=1}^{\infty}\dfrac{1}{n^{k}}&=\sum_{n=1}^{\infty}1\\&=\lim_{n\to \infty}\,\sum_{m=1}^{n}1\\&=\lim_{n\to \infty}n\\&=+\infty\end{aligned}}\) .

Let \(\displaystyle{k\in\left(0,1\right]}\) .

We define \(\displaystyle{f:\left[1,+\infty\right)\longrightarrow \mathbb{R}}\) by \(\displaystyle{f(x)=x^{-k}}\) .

The function \(\displaystyle{f}\) is continuous, positive, differentiable and strictly decreasing at \(\displaystyle{\left[1,+\infty\right)}\)

(cause \(\displaystyle{f^\prime(x)=-k\,x^{-k-1}<0\,,x\geq 1}\)). So, according to Cauchy's criterion, the series

\(\displaystyle{\sum_{n=1}^{\infty}f(n)=\sum_{n=1}^{\infty}\dfrac{1}{n^{k}}}\)

and the integral \(\displaystyle{\int_{1}^{\infty}f(x)\,\mathrm{d}x}\) have the same behavior. However,

\(\displaystyle{\bullet\,\,k=1}\) :

\(\displaystyle{\int_{1}^{\infty}f(x)\,\mathrm{d}x=\int_{1}^{\infty}\dfrac{1}{x}\,\mathrm{d}x=\left[\ln\,x\right]_{1}^{\infty}=\infty}\)

\(\displaystyle{\bullet\,\,k\in\left(0,1\right)}\) :

\(\displaystyle{\int_{1}^{\infty}f(x)\,\mathrm{d}x=\int_{1}^{\infty}x^{-k}\,\mathrm{d}x=\left[\dfrac{x^{1-k}}{1-k}\right]_{1}^{\infty}=\infty}\)

cause \(\displaystyle{1-k>0}\) .

and for this reason, the series \(\displaystyle{\sum_{n=1}^{\infty}\dfrac{1}{n^{k}}\,,k\in\left[0,1\right]}\) does not converge.
Post Reply

Create an account or sign in to join the discussion

You need to be a member in order to post a reply

Create an account

Not a member? register to join our community
Members can start their own topics & subscribe to topics
It’s free and only takes a minute

Register

Sign in

Who is online

Users browsing this forum: No registered users and 10 guests