Maths Olympiad Prep

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Algebra Difficulty 6.7 National olympiad Prove it

Lemma 2 Let y⩾2y \geqslant 2. We have
ln⁡ln⁡([y]+1)−ln⁡ln⁡2<∑2⩽k⩽y1kln⁡k<ln⁡ln⁡[y]+12ln⁡2−ln⁡ln⁡2\begin{array}{c} \ln \ln ([y]+1)-\ln \ln 2<\sum_{2 \leqslant k \leqslant y} \frac{1}{k \ln k} \\ \quad<\ln \ln [y]+\frac{1}{2 \ln 2}-\ln \ln 2 \end{array}

and
[y]{ln⁡[y]−1}+1<∑1⩽k⩽yln⁡k<([y]+1){ln⁡([y]+1)−1}+2−2ln⁡2\begin{array}{l} {[y]\{\ln [y]-1\}+1<\sum_{1 \leqslant k \leqslant y} \ln k} \\ \quad<([y]+1)\{\ln ([y]+1)-1\}+2-2 \ln 2 \end{array}

Solution

Prove that we have
\int_{k}^{k+1} \frac{\mathrm{d} t}{t \ln t}\int_{2}^{[y]+1} \frac{\mathrm{d} t}{t \ln t}=\ln \ln ([y]+1)-\ln \ln 2 . \end{array}

From the above two equations, we obtain equation (15). Similarly, from
\int_{k-1}^{k} \ln t \mathrm{~d} t\int_{1}^{[y]} \ln t \mathrm{~d} t=[y] \ln [y]-[y]+1 \end{array}

This proves equation (16).

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.