Prove that $\sum_{n=1}^N \ln(n)- \int_{0}^N \ln x +o(1) \geq 0.5*\ln(N)$

Question

Question: Prove that $\sum_{n=1}^N \ln(n)- \int_{0}^N \ln x +o(1) \geq 0.5*\ln(N)$, where 0(1) is infinity small and N is the last number of sum. My thoughts: In the graphical representation of the difference of integral and sum there should be triangles, whose squares must be equal to this value. Also there is no use of first element of the sum, because when n=0 -> lnn=0

Answer 1

Note: I added an upper bound to go with the lower bound.

I also added a proof that the series for $\ln(1+x)$ is enveloping.

I will show by elementary means that $\frac12\ln(N)+\frac32 \gt \sum_{n=1}^N\ln(n)-\int_{0}^N \ln(x) dx \gt \frac12\ln(N)+\frac13 $.

This shows that $\ln(N!) =N\ln(N)-N+\frac12\ln(N) +c(N) $ where $\frac13 \lt c(N) \lt \frac32 $. This agrees with Stirling's $\ln(\sqrt{2\pi}) \approx 0.9189$.

$\begin{array}\\ \ln(n)-\int_{n-1}^n \ln(x) dx &=\ln(n)-(x\ln(x)-x)_{n-1}^n\\ &=\ln(n)-(n\ln(n)-n-((n-1)\ln(n-1)-(n-1)))\\ &=\ln(n)-(n\ln(n)-1-((n-1)\ln(n-1)))\\ &=1-((n-1)\ln(n)-((n-1)\ln(n-1)))\\ &=1-(n-1)(\ln(n)-\ln(n-1))\\ &=1-(n-1)\ln(\frac{n}{n-1})\\ &=1-(n-1)\ln(1+\frac{1}{n-1})\\ \end{array} $

$\frac1{x}\ln(1+x) =\frac1{x}(x-\frac{x^2}{2}+\frac{x^3}{3}-...) =1-\frac{x}{2}+\frac{x^2}{3}-... $.

Since this series is enveloping (see the discussion at the bottom), $1-\frac{x}{2} \lt \frac1{x}\ln(1+x) \lt 1-\frac{x}{2}+\frac{x^2}{3} $ or $\frac{x}{2} \gt 1-\frac1{x}\ln(1+x) \gt \frac{x}{2}-\frac{x^2}{3} $.

Putting $x = \frac1{n-1}$, $\frac{1}{2(n-1)} \gt 1-(n-1)\ln(1+\frac1{n-1}) \gt \frac{1}{2(n-1)}-\frac{1}{3(n-1)^2} $.

Therefore

$\begin{array}\\ \sum_{n=2}^N\ln(n)-\int_{1}^N \ln(x) dx &=\sum_{n=2}^N(\ln(n)-\int_{n-1}^n \ln(x) dx)\\ &=\sum_{n=2}^N(1-(n-1)\ln(1+\frac{1}{n-1}))\\ &>\sum_{n=2}^N(\frac{1}{2(n-1)}-\frac{1}{3(n-1)^2})\\ &=\sum_{n=2}^N\frac{1}{2(n-1)}-\sum_{n=2}^N\frac{1}{3(n-1)^2}\\ &=\frac12\sum_{n=1}^{N-1}\frac{1}{n}-\frac13\sum_{n=1}^{N-1}\frac{1}{n^2}\\ \text{and}\\ \sum_{n=2}^N\ln(n)-\int_{1}^N \ln(x) dx &=\sum_{n=2}^N(1-(n-1)\ln(1+\frac{1}{n-1}))\\ &<\sum_{n=2}^N\frac{1}{2(n-1)}\\ &=\frac12\sum_{n=1}^{N-1}\frac{1}{n}\\ \end{array} $

Since $\frac1{x}$ is decreasing for $x > 0$, $\frac1{n} \gt \int_n^{n+1} \frac{dx}{x} $ so

$\begin{array}\\ \sum_{n=1}^{N-1}\frac{1}{n} &\gt \sum_{n=1}^{N-1} \int_n^{n+1} \frac{dx}{x}\\ &=\int_1^N \frac{dx}{x}\\ &=\ln(N)\\ \end{array} $

Similarly, $\frac1{n} \lt \int_{n-1}^{n} \frac{dx}{x} $ so

$\begin{array}\\ \sum_{n=1}^{N-1}\frac{1}{n} &=1+\sum_{n=2}^{N-1}\frac{1}{n}\\ &\lt 1+\sum_{n=2}^{N-1} \int_{n-1}^{n} \frac{dx}{x}\\ &=1+\int_1^{N-1} \frac{dx}{x}\\ &=1+\ln(N-1)\\ &<1+\ln(N)\\ \end{array} $

Also

$\begin{array}\\ \sum_{n=1}^{N-1}\frac{1}{n^2} &=1+\sum_{n=2}^{N-1}\frac{1}{n^2}\\ &=1+\sum_{n=2}^{N-1}\frac{1}{n^2}\\ &<1+\sum_{n=2}^{N-1}\frac{1}{n(n-1)}\\ &=1+\sum_{n=2}^{N-1}(\frac1{n-1}-\frac1{n})\\ &=1+1-\frac1{N-1}\\ &< 2\\ \end{array} $

so that $\sum_{n=2}^N\ln(n)-\int_{1}^N \ln(x) dx \gt \frac12\ln(N)-\frac23 $.

Since $\ln(1) = 0$ and $\int_0^1 \ln(x) dx =(x\ln(x)-x)_0^1 =-1 $, $\sum_{n=1}^N\ln(n)-\int_{0}^N \ln(x) dx \gt \frac12\ln(N)+\frac13 $.

Similarly, $\sum_{n=2}^N\ln(n)-\int_{1}^N \ln(x) dx \lt \frac12(1+\ln(N)) =\frac12+\frac12\ln(N) $ so that $\sum_{n=1}^N\ln(n)-\int_{0}^N \ln(x) dx \lt \frac12\ln(N)+\frac32 $.

On $\ln(1+x)$.

If $0 < x < 1$ and $m \in \mathbb{N}$,

$\begin{array}\\ \dfrac1{1+x} &=\sum_{n=0}^{\infty} (-1)^n x^n\\ &=\sum_{n=0}^{m-1} (-1)^n x^n+\sum_{n=m}^{\infty} (-1)^n x^n\\ &=\sum_{n=0}^{m-1} (-1)^n x^n+(-1)^mx^m\sum_{n=m}^{\infty} (-1)^{n-m} x^{n-m}\\ &=\sum_{n=0}^{m-1} (-1)^n x^n+(-1)^mx^m\sum_{n=0}^{\infty} (-1)^{n} x^{n}\\ &=\sum_{n=0}^{m-1} (-1)^n x^n+(-1)^mx^m\dfrac1{1+x}\\ \int_0^t \dfrac{dx}{1+x} &=\int_0^t\sum_{n=0}^{m-1} (-1)^n x^n+\int_0^t(-1)^mx^m\dfrac1{1+x}\\ &=\sum_{n=0}^{m-1} (-1)^n \int_0^t x^ndx+(-1)^m\int_0^t\dfrac{x^mdx}{1+x}\\ \ln(1+t) &=\sum_{n=0}^{m-1} \dfrac{(-1)^nt^{n+1}}{n+1}+(-1)^m\int_0^t\dfrac{x^mdx}{1+x}\\ &=\sum_{n=0}^{m-1} \dfrac{(-1)^nt^{n+1}}{n+1}+(-1)^mE_m(t)\\ \end{array} $

$\begin{array}\\ E_m(t) &=\int_0^t\dfrac{x^mdx}{1+x}\\ &\lt\int_0^tx^mdx\\ &=\dfrac{t^{m+1}}{m+1}\\ \text{and}\\ E_m(t) &=\int_0^t\dfrac{x^mdx}{1+x}\\ &\gt\int_0^t\dfrac{x^mdx}{1+t}\\ &=\dfrac{t^{m+1}}{(1+t)(m+1)}\\ \text{so}\\ 0 &\lt \dfrac{t^{m+1}}{m+1}-E_m(t)\\ &\lt \dfrac{t^{m+1}}{m+1}-\dfrac{t^{m+1}}{(1+t)(m+1)}\\ &= \dfrac{t^{m+1}}{m+1}(1-\dfrac{1}{1+t})\\ &= \dfrac{t^{m+1}}{m+1}(\dfrac{t}{1+t})\\ &= \dfrac{t^{m+2}}{(m+1)(1+t)}\\ &< \dfrac{t^{m+2}}{m+1}\\ \end{array} $