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I'd like to prove, that:


Of course it is smaller than $1$, but does the limit equal $1$?

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L'Hopital's rule, also. – Antonio Vargas Jun 17 '14 at 14:41
up vote 5 down vote accepted

Integrate by parts: $$\int_1^n\log t\mathrm dt=[t\log t]_{t=1}^{n}-\int_1^nt(\log t)'\mathrm dt=n\log n - n + 1\ .$$

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One can estimate rather than compute, that is, use the simple upper and lower bounds, valid for every positive $\varepsilon$, $$ n\log n\geqslant\int_1^n\log t\,\mathrm dt\geqslant\int_{\varepsilon n}^n\log t\,\mathrm dt\geqslant(1-\varepsilon)n\log(\varepsilon n)=(1-\varepsilon)n\log n+O(n). $$ This is helpful to see that the result is robust in the sense that it applies (with a little more work in the same vein) to every function $f$ such that $f(t)\sim\log t$ when $t\to+\infty$, even if $f$ has no handy explicit primitive.

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