Show that $limlimits_{Nrightarrowinfty}sumlimits_{n=1}^Nfrac{1}{N+n}=intlimits_1^2 frac{dx}{x}=ln(2)$

Question

Show that:
$$limlimits_{Nrightarrowinfty}sumlimits_{n=1}^Nfrac{1}{N+n}=intlimits_1^2 frac{dx}{x}=ln(2)$$

My attempt:
We build a Riemann sum with:
$1=x_0<x_1<...<x_{N-1}<x_N=2$
$x_n:=frac{n}{N}+1,,,,ninmathbb{N}_0$
That gives us:
$$sumlimits_{n=1}^N(x_n-x_{n-1})frac{1}{x_n}=sumlimits_{n=1}^N left(frac{n}{N}+1-left(frac{n-1}{N}+1right)right)frac{1}{frac{n}{N}+1}=sumlimits_{n=1}^N frac{1}{N}frac{N}{N+n}=sumlimits_{n=1}^Nfrac{1}{N+n}$$
We know from the definition, that:
$$limlimits_{Nrightarrowinfty}sumlimits_{n=1}^Nfrac{1}{N+n}=limlimits_{Nrightarrowinfty}sumlimits_{n=1}^N(x_n-x_{n-1})frac{1}{x_n}=intlimits_1^2 frac{dx}{x}$$
Now we show that,
$$intlimits_1^2 frac{dx}{x}=ln(2)$$
First we choose another Rieman sum with:
$1=x_0<x_1<...<x_{N-1}<x_N=2$
$x_n:=2^{frac{n}{N}},,,,ninmathbb{N}_0$
We get:
$$sumlimits_{n=1}^N(x_n-x_{n-1})frac{1}{x_n}=sumlimits_{n=1}^Nleft(2^{frac{n}{N}}-2^{frac{n-1}{N}}right)frac{1}{2^{frac{n-1}{N}}}=sumlimits_{n=1}^N 2^{frac{1}{N}}-1=Nleft(2^{frac{1}{N}}-1right)$$
Since we know that (with $x in mathbb{R})$:
$$limlimits_{xrightarrow0}frac{2^x-1}{x}=ln(2)Longrightarrow limlimits_{xrightarrow infty}x(2^{frac{1}{x}}-1)=ln(2)Longrightarrow limlimits_{Nrightarrow infty}N(2^{frac{1}{N}}-1)=ln(2)$$
We get:
$$ln(2)=limlimits_{Nrightarrow infty}N(2^{frac{1}{N}}-1)=limlimits_{Nrightarrow infty}sumlimits_{n=1}^Nleft(2^{frac{n}{N}}-2^{frac{n-1}{N}}right)frac{1}{2^{frac{n-1}{N}}}=intlimits_1^2 frac{dx}{x}=limlimits_{Nrightarrowinfty}sumlimits_{n=1}^Nfrac{1}{N+n}$$
$Box$

Hey it would be great, if someone could check my reasoning (if its correct) and give me feedback and tips :)

Jean L. · Answer

Your solution using a Riemann sum approximation to the integral $int^2_1frac{dx}{x}$ looks fine to me. Yves Daoust is much more direct.
A similar method was develop here to estimate another nice integral, I hope you appreciate it.
Here is a different method similar to Claude Leibovici's solution, but using a more elementary asymptotics for the harmonic sequence $H_n=sum^n_{k=1}frac{1}{k}$.
It is known that
$$
begin{align}
0<H_n-ln(n)-gamma < frac{1}{n+1}tag{1}label{one}
end{align}
$$
for all $ninmathbb{N}$, where  $gamma$ is a famous Euler-Mascheroni constant. The derivation of this is not difficult. It is based on a comparison between the integral $int^n_1frac{dx}{x}$ and $H_n$.
Using  $eqref{one}$ with $n=2N$ and $n=N$ gives
$$
0<H_{2N}-ln(2N)-gamma < frac{1}{2N+1}tag{2}label{two}
$$
$$
begin{align}
0<H_{N}-ln(N)-gamma < frac{1}{N+1}tag{3}label{three}
end{align}
$$
Subtracting $eqref{three}$ from $eqref{two}$ gives
$$
-frac{1}{N+1}< H_{2N}-H_N -ln(2N)+ln(N)<frac{1}{2N+1}
$$
The term $ln(2N)-ln(N)=ln(2)=int^2_1 frac{dx}{x}$. Then applying the squeeze lemma you obtain
$$
lim_{Nrightarrowinfty}sum^N_{n=1}frac{1}{N+n}=lim_{Nrightarrowinfty}sum^{2N}_{n=N+1}frac{1}{n}=lim_{Nrightarrowinfty}big(H_{2N}-H_Nbig)=ln 2
$$
I learned this method from this source where they use it to estimate another cool limit: $lim_{nrightarrowinfty}(H_{F_n}-H_{F_{n-1}} )$, there $F_n$ is the Fibonacci sequence.

user65203 · Answer

Your reasoning is correct, but you make it more complicated than needed.
$$frac1Nsum_{i=1}^Nfrac1{1+dfrac nN}toint_0^1frac{dx}{1+x}=left.ln(1+x)right|_0^1.$$

Claude Leibovici · Answer

Without Rieman sums.
$$S_N=sumlimits_{n=1}^Nfrac{1}{N+n}=H_{2 N}-H_N$$ Using the asymptotics of harmonic numbers
$$S_N=log (2)-frac{1}{4 N}+frac{1}{16 N^2}+Oleft(frac{1}{N^4}right)$$

Show that $limlimits_{Nrightarrowinfty}sumlimits_{n=1}^Nfrac{1}{N+n}=intlimits_1^2 frac{dx}{x}=ln(2)$

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