$\sum_{n=1}^{\infty}\frac{n^2}{a^2_{1}+a^2_{2}+\cdots+a^2_{n}}$is also convergent? Let sequence $a_{n}>0$, $n\in N^{+}$, and such
$\displaystyle\sum_{n=1}^{\infty}\dfrac{1}{a_{n}}$ convergent. Show that
$$\sum_{n=1}^{\infty}\dfrac{n^2}{a^2_{1}+a^2_{2}+\cdots+a^2_{n}}$$ is also convergent?
Jack A related result: maybe I guess this also is hold?
$$\sum_{k=1}^{n}\dfrac{k^2}{a^2_{1}+\cdots+a^2_{k}}\le\left(\dfrac{1}{a_{1}}+\cdots+\dfrac{1}{a_{n}}\right)^2$$

The Polya-Knopp's inequality (that is an instance of Hardy's inequality for negative exponents) states that for any $p\geq 1$ and for every positive sequence $\{a_n\}_{n\in\mathbb{N}}$ we have:
$$ \frac{N^{\frac{p+1}{p}}}{(p+1)\left(a_1^p+\ldots+a_N^p\right)^{1/p}}+\sum_{n=1}^N \left(\frac{n}{a_1^p+\ldots+a_n^p}\right)^{1/p} \leq (1+p)^{\frac{1}{p}}\sum_{n=1}^{N}\frac{1}{a_n},\tag{1}$$
hence by taking $p=2$ it follows that:
$$ \sum_{n= 1}^{N}\frac{\sqrt{n}}{\sqrt{a_1^2+a_2^2+\ldots+a_n^2}}\leq \sqrt{3}\sum_{n=1}^{N}\frac{1}{a_n}\tag{2} $$
Now we re-write the LHS of $(2)$ by partial summation. 
Let $Q_n^2\triangleq a_1^2+\ldots+a_n^2$ and $h(n)\triangleq\sum_{k=1}^{n}\sqrt{k}$:
$$\sum_{n=1}^N \frac{\sqrt{n}}{Q_n}=\frac{h(N)}{Q_N}-\sum_{n=1}^{N-1}h(n)\left(\frac{1}{Q_{n+1}}-\frac{1}{Q_n}\right)=\frac{h(N)}{Q_N}+\sum_{n=1}^{N-1}h(n)\frac{a_{n+1}^2}{Q_n Q_{n+1}(Q_{n+1}+Q_n)} $$
since $h(n)\geq\frac{2}{3}n^{3/2}$, it follows that:
$$ \frac{2N\sqrt{N}}{Q_N}+\sum_{n=1}^{N-1}\frac{n^{3/2} a_{n+1}^2}{Q_{n+1}^3}\leq 3\sqrt{3}\sum_{n=1}^{N}\frac{1}{a_n}.\tag{3}$$
If we let $g(n)=\sum_{k=1}^{n}k^2$ and apply partial summation to the original series we get:
$$ \sum_{n=1}^{N}\frac{n^2}{Q_n^2}=\frac{g(N)}{Q_N^2}+\sum_{n=1}^{N-1}g(n)\frac{a_{n+1}^2}{Q_n^2 Q_{n+1}^2}\tag{4}$$
hence by $(3)$ we just need to show that $\frac{g(n)}{Q_n Q_{n+1}}$ is bounded by some constant times $\frac{h(n)}{Q_n+Q_{n+1}}$, or:
$$ g(n)\left(Q_n+ Q_{n+1}\right) \leq K \cdot h(n) Q_n Q_{n+1} $$ 
or:
$$ \frac{1}{Q_n}+\frac{1}{Q_{n+1}}\leq K\cdot\frac{h(n)}{g(n)} \tag{5}$$
that follows from the fact that $\frac{\sqrt{n}}{Q_n}$ is summable by $(2)$.
Edit: A massive shortcut. If a positive sequence $\{b_n\}$ is such that $\sum {b_n}$ is convergent, then $\sum n b_n^2 $ is convergent too, since $\{n b_n\}$ must be bounded in order that $\sum b_n$ converges. So we can just use this lemma and $(2)$ to prove our claim.
