Show that:

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

My attempt:

We build a Riemann sum with:



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:



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:

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


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

  • 1
    $\begingroup$ Looks great to me! $\endgroup$
    – Tom Chen
    Jul 23 '20 at 2:18
  • 1
    $\begingroup$ No need to show $\int^2_1\frac1x\,dx=\log(2)$, that as $x\rightarrow\int^x_1\frac{1}{t}\,dt$ is, by definition, $\log x$. $\endgroup$ Jul 23 '20 at 12:25
  • $\begingroup$ yeah, well in my book it wasnt clear. The exercise before I had to show: $\int_1^a \frac{dx}{x}=\ln(a)$ I put the way I solved that into this post, so anyone could see it :) $\endgroup$ Jul 23 '20 at 12:34

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

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

  • $\begingroup$ Yeah maybe I should write down restrictions I had in the future. It was the chapter before differentiation and integration got related. The chapter tried to bring the idea closer with riemann sums. Only by knowledge about series and limits I had to show it. But sometimes its also interesting to see how other people solve it with different ideas. Ofc the part with showing that the integral is a certain value is completly trivial without restrictions :) $\endgroup$ Jul 23 '20 at 12:49
  • $\begingroup$ There should be a limit $N\to\infty$, but I fully agree that this is the better way to go $\endgroup$
    – LL 3.14
    Jul 23 '20 at 19:44
  • $\begingroup$ @LL3.14: yes, of course. $\endgroup$
    – user65203
    Jul 23 '20 at 21:14
  • $\begingroup$ @CoffeeArabica: restrictions ? $\endgroup$
    – user65203
    Jul 23 '20 at 21:15

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.


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 $$S_N=\log (2)-\frac{1}{4 N}+\frac{1}{16 N^2}+O\left(\frac{1}{N^4}\right)$$


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