Prove that $A$ is bounded operator on $\ell^p$ and find $\| A\|$ 
For which one $p \ge 1$ is with $$A(x_n)_{n=1}^{\infty}=\left(\frac{1}{m}\sum_{n=1}^{m}\frac{x_n}{\sqrt{n}}\right)_{m=1}^{\infty}$$ defined bounded linear operator $A:\ell^p \to \ell^p$? Find norm $\| A\|.$

First, I have to show that if $x = (x_n)_{n=1}^{\infty} \in \ell^p$ then $\displaystyle y=(y_m)_{m=1}^{\infty} \stackrel{\text{def}}{=}\left(\frac{1}{m}\sum_{n=1}^{m}\frac{x_n}{\sqrt{n}}\right)_{m=1}^{\infty} \in \ell^p$, that is $\displaystyle\sqrt[p]{\sum_{m=1}^{\infty}|y_m|^p} < +\infty.$
All right, $$\begin{align*} |y_m|=\left|\frac{1}{m}\sum_{n=1}^{m}\frac{x_n}{\sqrt{n}}\right|&\leq\frac{1}{m} \sum_{n=1}^{m} \frac{|x_n|}{\sqrt{n}}\\ &\stackrel{\star}{\leq}\frac{1}{m} \sqrt[p]{\sum_{n=1}^{m}|x_n|^p} \sqrt[q]{\sum_{n=1}^{m} n^{-q/2}} \\
&\leq \frac{\| x \|_p}{m}\sqrt[q]{\sum_{n=1}^{m} n^{-q/2}}.\end{align*}$$
So $$\begin{align*} \sum_{m=1}^{\infty} |y_m|^p &\leq \sum_{m=1}^{\infty} \left|\frac{\| x \|_p}{m}\sqrt[q]{\sum_{n=1}^{m} n^{-q/2}}\right|^p\\&=\sum_{m=1}^{\infty} \frac{1}{m^p} \| x\|^p_p \left(\sum_{n=1}^{m} n^{-q/2}\right)^{p/q}\\ &\leq \|x\|^p_p \sum_{m=1}^{\infty} \frac{1}{m^p}   \left(\sum_{n=1}^{\infty} n^{-q/2}\right)^{p/q} \end{align*}$$
If I see well, last sum is convergent for $q/2 >1 \iff q>2$ and $p>1$ (but with condition $1/p + 1/q =1$).
I think my estimate is too sharp.
And we find that $$\| A \| \leq \sqrt[p]{\sum_{m=1}^{\infty} \frac{1}{m^p}   \left(\sum_{n=1}^{\infty} n^{-q/2}\right)^{p/q}}$$
But for part (if my approximation is good enough) with $\geq$ I don't have idea (to be honest, I used Hölder inequality, so we want $x_n = c  \frac{1}{\sqrt{n}}$, that will give us something divergent I think so).
$\star$ - one doubt also: I used here Hölder inequality, but that works only for $p,q >1$, but in my question we have $p \geq 1$, so basically I have to work case $p=1$ as separate? 
 A: As mention by robjohn, only the case $p>1$ allows $A$ to be bounded. 
In your last estimate, you changed $m$ to $\infty$, and there was no reason for that step, as you already had an upper bound. 
One way to get a lower bound for the norm of $A$ is as follows. Fix  $k\in\mathbb N$. Let
$$
x_n=n^{-q/2p}. 
$$
Then, since $-q/2p-1/2=-q/2$,
\begin{align}
\|Ax\|_p^p&=\sum_{m=1}^\infty|(Ax)_m|^p
=\sum_{m=1}^\infty \left|\frac1m\,\sum_{n=1}^m\frac{n^{-q/2p}}{n^{1/2} }\right|^p\\ \ \\
&=\sum_{m=1}^\infty \left(\frac1m\,\sum_{n=1}^m{n^{-q/2}}\right)^p\\ \ \\
&=\sum_{m=1}^\infty \frac1{m^p}\left[\left(\sum_{n=1}^m{|n^{-q/2p}|^p}\right)^{1/p}\left(\sum_{n=1}^m{n^{-q/2}}\right)^{1/q}\right]^p\\ \ \\
&=\|x\|_p^p\,\sum_{m=1}^\infty\frac1{m^p}\left(\sum_{n=1}^m{n^{-q/2}}\right)^{p/q}.
\end{align}
So 
$$
\|A\|\geq
\left(\sum_{m=1}^\infty\frac1{m^p}\,\left(\sum_{n=1}^mn^{-q/2}\right)^{p/q}\right)^{1/p},
$$
as you already had the upper bound. 
The series converges for any $p>1$. Indeed, for $m$ large enough we have 
$$
\sum_{n=1}^mn^{-q/2}\leq 1+\int_1^mt^{-q/2}=1+\frac{m^{1-q/2}-1}{1-q/2}\leq c\,m^{1/2}
$$
for some $c>0$. So
$$
\frac1{m^p}\,\left(\sum_{n=1}^mn^{-q/2}\right)^{p/q}\leq c^{1/q}\,m^{p/2q-p}
=c^{1/q}\,m^{-p/2-1/2}.
$$
Since $p>1$, the series $\sum_{m=1}^\infty\frac1{m^p}\,\left(\sum_{n=1}^mn^{-q/2}\right)^{p/q}$ converges. 
A: Brute force and Hölder give us, with $\frac1p+\frac1q=1$,
$$
\begin{align}
\sum_{m=1}^\infty\left|\frac1m\sum_{n=1}^m\frac{x_n}{\sqrt{n}}\right|^p
&\le\sum_{m=1}^\infty\frac1{m^p}\left(\sum_{n=1}^m\frac{|x_n|}{\sqrt{n}}\right)^p\\
&\le\sum_{m=1}^\infty\frac1{m^p}\|x_n\|_p^p\left(\sum_{n=1}^m\frac1{n^{q/2}}\right)^{p/q}\\
&\le\|x_n\|_p^p\sum_{m=1}^\infty\frac1{m^p}\left(2\sqrt{m}\right)^{p-1}\tag{$q\ge1$}\\
&=2^{p-1}\|x_n\|_p^p\zeta\left(\frac{p+1}{2}\right)
\end{align}
$$
Thus, $A$ is bounded for $p\gt1$ with norm no greater than $2^{1-1/p}\zeta\left(\frac{p+1}{2}\right)^{1/p}\le2\,\zeta\left(\frac{p+1}{2}\right)$.
For $(x_n)=(1,0,0,\dots)\in\ell^1$,
$$
\frac1m\sum_{n=1}^m\frac{x_n}{\sqrt{n}}=\frac1m
$$
Thus, we cannot have that $A$ is bounded on $\ell^1$.
