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Prove that $n\log(n)-n+1 \leq \log(n!) \leq (n+1)\log(n+1)-(n+1)-(2\log(2)-2)$.

Approximating $\int_{1}^{n}\log(x)dx$ using a Riemann sum at right endpoints is $\log(2) + \log(3)+\dots +\log(n) = \log(n!)$ which overestimates so the claim for the lower bound follows. For the upper bound we integrate $\int_{2}^{n+1}\log(x)dx$. I can prove the upper bound if I can show that $\log(1+\frac{1}{n}) \leq \frac{1}{n}$. I know that the Taylor expansion of $\log(1+\frac{1}{n})$ is an alternating series but I am not sure how to show this.

I hope someone can provide some guidance and also clarify why is $\int_{1}^{n}\log(x)dx \leq \int_{2}^{n+1}\log(x)dx$. Is this because $\log(x)$ is an increasing function?

Another User
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rachelhoward
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1 Answers1

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The fact that $\log(1+x) \leq x$ for $x>-1$ is a consequence of just concavity, no higher than second derivatives are required. This is a general principle: concave functions lie below their tangent lines.

In general for an increasing function $f$ and a fixed $b>0$, $F(a)=\int_a^{a+b} f(x) dx$ is an increasing function of $a$ (when it makes sense). Thus in your case it is important that the lengths of the two intervals of integration are the same.

Ian
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