Show that $\lim\limits_{N\rightarrow\infty}\sum\limits_{n=1}^N\frac{1}{N+n}=\int\limits_1^2 \frac{dx}{x}=\ln(2)$

Question

Show that:

$$\lim\limits_{N\rightarrow\infty}\sum\limits_{n=1}^N\frac{1}{N+n}=\int\limits_1^2 \frac{dx}{x}=\ln(2)$$

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My attempt:

We build a Riemann sum with:

$1=x_0<x_1<...<x_{N-1}<x_N=2$

$x_n:=\frac{n}{N}+1,\,\,\,n\in\mathbb{N}_0$

That gives us:

$$\sum\limits_{n=1}^N(x_n-x_{n-1})\frac{1}{x_n}=\sum\limits_{n=1}^N \left(\frac{n}{N}+1-\left(\frac{n-1}{N}+1\right)\right)\frac{1}{\frac{n}{N}+1}=\sum\limits_{n=1}^N \frac{1}{N}\frac{N}{N+n}=\sum\limits_{n=1}^N\frac{1}{N+n}$$

We know from the definition, that:

$$\lim\limits_{N\rightarrow\infty}\sum\limits_{n=1}^N\frac{1}{N+n}=\lim\limits_{N\rightarrow\infty}\sum\limits_{n=1}^N(x_n-x_{n-1})\frac{1}{x_n}=\int\limits_1^2 \frac{dx}{x}$$

Now we show that,

$$\int\limits_1^2 \frac{dx}{x}=\ln(2)$$

First we choose another Rieman sum with:

$1=x_0<x_1<...<x_{N-1}<x_N=2$

$x_n:=2^{\frac{n}{N}},\,\,\,n\in\mathbb{N}_0$

We get:

$$\sum\limits_{n=1}^N(x_n-x_{n-1})\frac{1}{x_n}=\sum\limits_{n=1}^N\left(2^{\frac{n}{N}}-2^{\frac{n-1}{N}}\right)\frac{1}{2^{\frac{n-1}{N}}}=\sum\limits_{n=1}^N 2^{\frac{1}{N}}-1=N\left(2^{\frac{1}{N}}-1\right)$$

Since we know that (with $x \in \mathbb{R})$:

$$\lim\limits_{x\rightarrow0}\frac{2^x-1}{x}=\ln(2)\Longrightarrow \lim\limits_{x\rightarrow \infty}x(2^{\frac{1}{x}}-1)=\ln(2)\Longrightarrow \lim\limits_{N\rightarrow \infty}N(2^{\frac{1}{N}}-1)=\ln(2)$$

We get:

$$\ln(2)=\lim\limits_{N\rightarrow \infty}N(2^{\frac{1}{N}}-1)=\lim\limits_{N\rightarrow \infty}\sum\limits_{n=1}^N\left(2^{\frac{n}{N}}-2^{\frac{n-1}{N}}\right)\frac{1}{2^{\frac{n-1}{N}}}=\int\limits_1^2 \frac{dx}{x}=\lim\limits_{N\rightarrow\infty}\sum\limits_{n=1}^N\frac{1}{N+n}$$

$\Box$

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

Answer 1

Without Rieman sums. $$S_N=\sum\limits_{n=1}^N\frac{1}{N+n}=H_{2 N}-H_N$$ Using the asymptotics of harmonic numbers $$S_N=\log (2)-\frac{1}{4 N}+\frac{1}{16 N^2}+O\left(\frac{1}{N^4}\right)$$

Answer 2

Your reasoning is correct, but you make it more complicated than needed.

$$\frac1N\sum_{i=1}^N\frac1{1+\dfrac nN}\to\int_0^1\frac{dx}{1+x}=\left.\ln(1+x)\right|_0^1.$$

Answer 3

Your solution using a Riemann sum approximation to the integral $\int^2_1\frac{dx}{x}$ 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 $n\in\mathbb{N}$, where $\gamma$ is a famous Euler-Mascheroni constant. The derivation of this is not difficult. It is based on a comparison between the integral $\int^n_1\frac{dx}{x}$ 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 $$ \lim_{N\rightarrow\infty}\sum^N_{n=1}\frac{1}{N+n}=\lim_{N\rightarrow\infty}\sum^{2N}_{n=N+1}\frac{1}{n}=\lim_{N\rightarrow\infty}\big(H_{2N}-H_N\big)=\ln 2 $$

I learned this method from this source where they use it to estimate another cool limit: $\lim_{n\rightarrow\infty}(H_{F_n}-H_{F_{n-1}} )$, there $F_n$ is the Fibonacci sequence.