Question Number 219936 by Razafitiana last updated on 04/May/25
$$\mathrm{Prove}\:\mathrm{that}:\forall\mathrm{n}\in\mathrm{IN} \\ $$$$\underset{\:\mathrm{n}} {\int}^{\:\mathrm{n}+\mathrm{1}} \mathrm{ln}\left(\mathrm{t}\right)\mathrm{dt}\leqslant\mathrm{ln}\left(\mathrm{n}+\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$
Answered by MrGaster last updated on 04/May/25
![∫^( n+1) _( n) ln(t)dt=ln[t ln t−t]_n ^(n+1) =(n+1)n(n+1)−(n+1)−(n ln n−n) =(n+1)ln(n+1)−n ln n−1 ln(n+(1/2))(false) =(n+1)ln(((n+1)/n))−1 [Q.E.D]](https://www.tinkutara.com/question/Q219947.png)
$$\underset{\:\mathrm{n}} {\int}^{\:\mathrm{n}+\mathrm{1}} \mathrm{ln}\left(\mathrm{t}\right)\mathrm{dt}=\mathrm{ln}\left[{t}\:\mathrm{ln}\:{t}−{t}\right]_{{n}} ^{{n}+\mathrm{1}} \\ $$$$=\left({n}+\mathrm{1}\right){n}\left({n}+\mathrm{1}\right)−\left({n}+\mathrm{1}\right)−\left({n}\:\mathrm{ln}\:{n}−{n}\right) \\ $$$$=\left({n}+\mathrm{1}\right)\mathrm{ln}\left({n}+\mathrm{1}\right)−{n}\:\mathrm{ln}\:{n}−\mathrm{1} \\ $$$$\cancel{\mathrm{ln}\left({n}+\frac{\mathrm{1}}{\mathrm{2}}\right)}\left(\mathrm{false}\right) \\ $$$$=\left({n}+\mathrm{1}\right)\mathrm{ln}\left(\frac{{n}+\mathrm{1}}{{n}}\right)−\mathrm{1} \\ $$$$\left[\mathrm{Q}.\mathrm{E}.\mathrm{D}\right] \\ $$