59. First prove
$$H_{k} \leqslant \frac{4}{k(k+1)^{2}}\left(a_{1}+4 a_{2}+\cdots+k^{2} a_{k}\right)$$

By the Cauchy-Schwarz inequality, we have
$$\begin{array}{l} \left(\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{k}}\right)\left(a_{1}+4 a_{2}+\cdots+k^{2} a_{k}\right) \geqslant \\ (1+2+\cdots+k)^{2}=\frac{k^{2}(k+1)^{2}}{4} \end{array}$$

Therefore,
$$\begin{array}{c} H_{k} \leqslant \frac{4}{k(k+1)^{2}}\left(a_{1}+4 a_{2}+\cdots+k^{2} a_{k}\right) \\ H_{1}+H_{2}+\cdots+H_{n} \leqslant \alpha_{1} a_{1}+\alpha_{2} a_{2}+\cdots+\alpha_{n} a_{n} \end{array}$$

Thus,
where
$$\alpha_{i}=4 i^{2}\left(\frac{1}{i(i+1)^{2}}+\frac{1}{(i+1)(i+2)^{2}}+\cdots+\frac{1}{n(n+1)^{2}}\right), i=1,2, \cdots, n$$

Also,
$$\frac{1}{m(m+1)^{2}}<\frac{1}{2} \cdot \frac{2 m+1}{m^{2}(m+1)^{2}}=\frac{1}{2}\left(\frac{1}{m^{2}}-\frac{1}{(m+1)^{2}}\right)$$

Therefore,
$$\begin{array}{l} \alpha_{i}<4 i^{2} \cdot \frac{1}{2}\left[\left(\frac{1}{i^{2}}-\frac{1}{(i+1)^{2}}\right)+\left(\frac{1}{(i+1)^{2}}-\right.\right. \\ \left.\left.\frac{1}{(i+2)^{2}}\right)+\cdots+\left(\frac{1}{n^{2}}-\frac{1}{(n+1)^{2}}\right)\right]< \\ 4 i^{2} \cdot \frac{1}{2}\left(\frac{1}{i^{2}}-\frac{1}{(n+1)^{2}}\right)<4 i^{2} \cdot \frac{1}{2 i^{2}}=2 \end{array}$$

Thus,
$$\begin{array}{l} H_{1}+H_{2}+\cdots+H_{n} \leqslant \alpha_{1} a_{1}+\alpha_{2} a_{2}+\cdots+\alpha_{n} a_{n}< \\ 2\left(a_{1}+a_{2}+\cdots+a_{n}\right)=2 \end{array}$$