Let $\{a_n\}$ be a sequnce of positive numbers satisfying $$a_{n+1}=a_n+\frac{a_n^{\alpha}}{n^{\alpha}}, \alpha>1, a_1\in (0,1),$$ prove that $\{a_n\}$ is bounded.
I tried this way:
$$ \frac{1}{a_{n+1}}=\frac{1}{a_n+\frac{a_n^{\alpha}}{n^\alpha}}=\frac{n^{\alpha}}{a_n n^{\alpha}+a_n^{\alpha}}=\frac{1}{a_n}\left(1-\frac{a_n^{\alpha-1}}{n^{\alpha}+a_n^{\alpha-1}}\right)=\frac{1}{a_n}-\frac{a_n^{\alpha-2}}{n^\alpha+a_n^{\alpha-1}}$$
$$ \Rightarrow \frac{1}{a_{n+1}}=\sum_{k=1}^{n}\left(\frac{1}{a_{k+1}}-\frac{1}{a_k}\right)+\frac{1}{a_1}= -\sum_{k=1}^{n}\frac{a_k^{\alpha-2}}{k^{\alpha}+a_k^{\alpha-1}}+\frac{1}{a_1}$$
Then I want to prove that this formula is greater than a certain positive number, but failed. I am not sure whether the way is effective. Can someone help me? Thanks.
