If (z_n) is bounded, prove that the following are also…

If $(z_n)$ is bounded, prove that the following are also bounded: $\ 1^\circ\ \left(1+\dfrac{z_n}n\right)^n;\quad2^\circ\ \dfrac{z_1+\cdots+z_n}n;\quad3^\circ\ \dfrac{z_1+2z_2+\cdots+nz_n}{n^2};\quad4^\circ\ \dfrac{p_1z_1+\cdots+p_nz_n}{p_1+\cdots+p_n}\ (p_i>0);\quad5^\circ\ \sqrt[n]{z_1z_2\cdots z_n}$.

Part $1^\circ$: $\left(1+\dfrac{z_n}n\right)^n$

$$ \left|1+\frac{z_n}n\right|\le1+\frac{|z_n|}n\le1+\frac Mn \qquad\Longrightarrow\qquad \left|\left(1+\frac{z_n}n\right)^n\right|\le\left(1+\frac Mn\right)^n $$

For real $M\ge0$, the sequence $\left(1+\frac Mn\right)^n$ is classically known to increase monotonically in $n$ toward its limit $e^M$, so $\left(1+\frac Mn\right)^n<e^M$ for every finite $n$. Hence:

$$ \boxed{\left|\left(1+\frac{z_n}n\right)^n\right|<e^M} $$

Part $2^\circ$: the Cesàro mean $\dfrac{z_1+\cdots+z_n}n$

$$ \left|\frac{z_1+\cdots+z_n}n\right|\le\frac{|z_1|+\cdots+|z_n|}n\le\frac{nM}n=M $$

Part $3^\circ$: $\dfrac{z_1+2z_2+\cdots+nz_n}{n^2}$

$$ \left|\frac{\sum_{k=1}^nkz_k}{n^2}\right|\le\frac{\sum_{k=1}^nk|z_k|}{n^2}\le\frac{M\sum_{k=1}^nk}{n^2}=\frac{M\cdot\frac{n(n+1)}2}{n^2}=\frac M2\left(1+\frac1n\right)\le M $$

(using $1+\frac1n\le2$ for $n\ge1$).

Part $4^\circ$: the weighted average $\dfrac{\sum p_kz_k}{\sum p_k}$, $p_k>0$

$$ \left|\frac{\sum_{k=1}^np_kz_k}{\sum_{k=1}^np_k}\right|\le\frac{\sum_{k=1}^np_k|z_k|}{\sum_{k=1}^np_k}\le\frac{M\sum_{k=1}^np_k}{\sum_{k=1}^np_k}=M $$

(this is the most general of the averaging constructions; part $2^\circ$ is the special case $p_k\equiv1$).

Part $5^\circ$: the geometric mean $\sqrt[n]{z_1z_2\cdots z_n}$

$$ \left|\sqrt[n]{z_1z_2\cdots z_n}\right|=\sqrt[n]{|z_1||z_2|\cdots|z_n|}\le\sqrt[n]{M^n}=M $$