The series does not converge
- Tolaso J Kos
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The series does not converge
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.
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.
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Re: The series does not converge
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.
\(\displaystyle{\begin{aligned} \sum_{n=1}^{\infty}\dfrac{1}{n^{k}}&=\sum_{n=1}^{\infty}1\\&=\lim_{n\to \infty}\,\sum_{m=1}^{n}1\\&=\lim_{n\to \infty}n\\&=+\infty\end{aligned}}\) .
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.
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