Ninth STARS OF MATHEMATICS Competition, 2015

Set $b_0 = 1$ and $b_k = 1 + a_1 + \dots + a_k$, $k = 1, \dots, n$, to obtain a strictly increasing string of positive integers $1 = b_0 < b_1 < \dots < b_n$, and write the sum in the left-hand member in the form $\sum_{k=1}^{n} (b_k - b_{k-1})^{1/2}/b_k$.

Next, let $m = \min\{k: b_k > n^2\} \ge 1$; if there is no $b_k > n^2$, let $m = n + 1$, to split the above sum into
$$\sum_{k=1}^{m-1} \frac{\sqrt{b_k - b_{k-1}}}{b_k} + \sum_{k=m}^{n} \frac{\sqrt{b_k - b_{k-1}}}{b_k}, \quad (*)$$
where empty sums are zero. We show that the first sum does not exceed $\sum_{k=2}^{n^2} 1/k$, and the second is always less than 1.

If $k = 1, \dots, m-1$, write $(b_k - b_{k-1})^{1/2}/b_k \le (b_k - b_{k-1})/b_k = 1/b_k + \dots + 1/b_k \le \sum_{j=b_{k-1}+1}^{b_k} 1/j$, to deduce that the first sum in $(*)$ does not exceed $\sum_{k=1}^{m-1} \sum_{j=b_{k-1}+1}^{b_k} 1/j = \sum_{k=2}^{b_{m-1}} 1/k \le \sum_{k=2}^{n^2} 1/k$.

Finally, if $k = m, \dots, n$, write $(b_k - b_{k-1})^{1/2}/b_k < b_k^{-1/2} < 1/n$, to deduce that the second sum in $(*)$, when non-empty, is less than $(n - m + 1)/n \le 1$. The conclusion follows.