2015 Ninth STARS OF MATHEMATICS Competition

Set $b_0=1$ and $b_k=1+a_1+\cdots +a_k$, $k=1,\dots ,n$, to obtain a strictly increasing string of positive integers $1=b_0<b_1<\cdots <b_n$, 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 $$\sum _{k=1}^{m-1}\frac{\sqrt{b_k-b_{k-1}}}{b_k}+\sum _{k=m}^n\frac{\sqrt{b_k-b_{k-1}}}{b_k},(\ast )$$ 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 $(b_k-b_{k-1}{)}^{1/2}/b_k\le (b_k-b_{k-1})/b_k=1/b_k+\cdots +1/b_k\le {\sum }_{j=b_{k-1}+1}^{b_k}1/j$, to deduce that the first sum in $(\ast )$ 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 $(\ast )$, when non-empty, is less than $(n-m+1)/n\le 1$. The conclusion follows.