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I'm interested in knowing whether there exists a closed form of $l^1$ and $l^2$ norms of the three following sequences (except for $u_n$ that doesn't admit a $l_1$ norm) and try to calculate the resulting norms.

$$u_n=\dfrac{n}{n^2+1} , v_n=\sin^2\frac{\pi}n,\\ w_n=\dfrac1{\sqrt{n^2+n}}$$

What I've tried:

For $u$, $v$ and $w$ I tried comparison with integral. The issue is that when comparing a sum $S_n$ I get:

$$\int_1^{N+1} f \leq S_n \leq \int_0^N f,$$

where $f$ is an associated function to each sequences (here decreasing functions) $u_n=f(n)$.

Because of the bound $1$ and $0$ which are different for those finite integrals I can't deduce a finite value for the norms I want.

I've also tried to develop the $\sin^2$ or to develop function in whole series without success.

If you are aware of the existence of a closed form of one of these functions, please let me know with a proof justifying the fact.

Anne Bauval
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EDX
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    The $\ l_2\ $ norm of $\ w\ $, at least, is easy, because $\ \displaystyle \sum_{n=1}^\infty w_n^2\ $ is a telescoping series: $$ w_n^2=\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}\ |w|2=\sqrt{\sum{n=1}^\infty w_n^2}\ =1\ . $$ – lonza leggiera Sep 14 '20 at 12:54
  • Oh yes, thanks! – EDX Sep 14 '20 at 19:11
  • The $l_2$ norm of $u$ can be calculated in the following way: $\sum_{n=1}^{\infty} (u_n)^2 = \sum_{n=1}^{\infty} (n + 1)^{-2} = \sum_{n=2}^{\infty} (n)^{-2} = -1 + \sum_{n=1}^{\infty} (n)^{-2} = \pi^2/6 - 1$. So, $\lVert u \rVert_2 = \sqrt{\pi^2/6 - 1}$. – jan May 11 '23 at 14:08
  • @jan $\sum_{n=1}^{\infty} (u_n)^2\ne \sum_{n=1}^{\infty} (n + 1)^{-2}.$ – Anne Bauval May 11 '23 at 15:07
  • @jan I think you misread the definition of $u_n$ – jjagmath May 11 '23 at 15:09

1 Answers1

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As you already mentionned, $\|u\|_1=+\infty.$

Similarly, $\|w\|_1=+\infty.$

$$\begin{align}\|u\|_2^2&=\sum_{n\ge1}\frac{n^2}{(n^2+1)^2}\\&=\sum_{n\ge1}\frac1{n^2+1}-\sum_{n\ge1}\frac1{(n^2+1)^2}\\&=\left(\frac\pi2\coth\pi-\frac12\right)-\left(-\frac12+\frac\pi4\coth\pi+\frac{\pi^2}{4\sinh^2\pi}\right)\\ &=\frac\pi4\left(\coth\pi-\frac\pi{\sinh^2\pi}\right), \end{align}$$ where the calculations of $\sum_{n\ge1}\frac1{n^2+1}$ and $\sum_{n\ge1}\frac1{(n^2+1)^2}$ can be found for instance in these posts: 1, 2, 3.

The last norm was calculated in @lonzaleggiera's comment, and can also be found in this post: 4: $$\|w\|_2^2=\sum_{n\ge1}\frac1{n\left(n+1\right)}=1.$$

There remains to calculate the $\ell^1$ and $\ell^2$ norm of $v.$

Anne Bauval
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