0

I'm currently stuck with the following question:

Prove, that $\ln(2) = \lim\limits_{n \to \infty} \sum_{k=1}^n \frac{1}{n+k}$ by rewriting the left side as an Integral.

So my current thoughts are:

$\ln(2) = \int_1^2 \frac{1}{x} \mathrm{d}x$

$\lim\limits_{n \to \infty} \sum_{k=1}^n \frac{1}{n+k} = \lim\limits_{n \to \infty} \sum_{k=1}^n \frac{1}{n} \frac{1}{\frac{k}{n} + 1} = \lim\limits_{n \to 0} \sum_{k=1}^{\frac{2-1}{n}} n \frac{1}{nk + 1}$

Like this, the sum is rewritten as a step function with the width of each equidistant step decreasing. If the sum is the upper sum, I have to show, that $\frac{1}{nk+1}=\ln(1 + \frac{k+1}{n})$. How can I "get rid" if the $\ln$?

Tim
  • 127
  • 6

2 Answers2

1

hint

Write it as

$$\frac{b-a}{n}\sum_{k=1}^nf\left(a+k\frac{b-a}{n}\right)$$

the limit will be $$\int_a^bf(x)dx$$

In your case, $a=0\; \; b=1, f(x)=\frac{1}{1+x}$ Or $a=1,\;b=2 \;$ and$ \; f(x)=\frac 1x.$

J.G.
  • 115,835
hamam_Abdallah
  • 62,951
  • So it would be $\int_1^2 \frac{1}{x} dx = \frac{1}{n} \sum_{k=1}^n \frac{1}{1 + \frac{k}{n}}$ which is equal to my upper transformation. But how can I prove the equivalence you provided to me? – Tim Feb 01 '19 at 17:29
1

You're almost there, just use the approximation of the integral by a Riemann sum over the grid $1, 1+ 1/n,\ldots, 2$, then

$$\lim\limits_{n \to \infty} \sum_{k=1}^n \frac{1}{n} \frac{1}{\frac{k}{n} + 1}= \lim\limits_{n \to \infty} \sum_{k=1}^n \left[\left(1+\frac{k}{n}\right)-\left(1+\frac{k-1}{n}\right)\right] \frac{1}{1+\frac{k}{n}}=\int_1^2\frac{1}{x}dx.$$

Mars Plastic
  • 4,239
  • So $\left[\left(1+\frac{k}{n}\right)-\left(1+\frac{k-1}{n}\right)\right]$ is the "width" of the subdivision and $\frac{1}{1+ \frac{k}{n}}$ is the "height" respectively. But where do I know from, that this infinite sum is equal to the Integral? – Tim Feb 01 '19 at 17:43
  • That's the definition of the Riemann integral. And be careful: It's not an infinite sum, but the limit of a sequence of finite sums. – Mars Plastic Feb 01 '19 at 17:48
  • Ok, thank you very much. We haven't defined the Riemann sum in our lecture yet, is there a way of solving the problem without this definition? – Tim Feb 01 '19 at 18:15
  • OK, so how is the integral introduced (defined) in your lecture? – Mars Plastic Feb 02 '19 at 01:20
  • We have introduced the integral with step functions: Let $\varphi$ be a step function, then the integral of $\varphi$ over $[a,b]$ is defined as $\int_a^b \varphi(x) dx = \sum_{i=1}^n c_i (x_i - x_{i-1})$. And we have the definition of the integral of a regulated function: If $(\varphi_n)$ is uniformly converging to $f$, then $\int_a^b f(x) dx = \lim\limits{n \to \infty}\int_a^b \varphi_n(x) dx$. – Tim Feb 02 '19 at 06:53
  • Alright, then you just have to find the right step function $\varphi_n$ such that

    $$\int_1^2\varphi_n(x)dx=\sum_{k=1}^n \left[\left(1+\frac{k}{n}\right)-\left(1+\frac{k-1}{n}\right)\right] \frac{1}{1+\frac{k}{n}}$$

    and $\varphi_n\to f$ uniformly. It works. ;-)

     – Mars Plastic Feb 02 '19 at 16:30
  • Sorry for asking so many questions and thank you very much for your patience. I don't understand how the term in the sum does uniformly converge to $f$, as there is no $x$ in the term. Or do I have to use something like $\left[ (1+\frac{k}{n} ) - (1 + \frac{k-1}{n}) \right] \frac{1}{x(1+\frac{k}{n})}$ in the sum? – Tim Feb 02 '19 at 17:47
  • No, this sum converges to the integral of $f$. Maybe look at the function

    $$\varphi_n\colon[1,2]\to\mathbb R,\quad x\mapsto\sum_{k=1}^n \chi_{\left[1+\frac{k-1}{n},1+\frac{k}{n}\right]}(x) \frac{1}{1+\frac{k}{n}},$$

    where $\chi_A$ denotes the characteristic function of the set $A$, i.e.

    $$\chi_A(x):=\begin{cases} 1, & x\in A, \ 0, & x\notin A.\end{cases}$$

    Is this $\varphi_n$ a step function? What's its integral from 1 to 2? Does $\varphi_n$ converge uniformly?

     – Mars Plastic Feb 02 '19 at 20:03
  • $\varphi_n$ is a step function, as for every $x \in \left(1+\frac{k-1}{n}, 1+\frac{k}{n}\right)$ the value of $\varphi_n$ is constant with $\varphi_n(x)=\frac{1}{1+\frac{k}{n}}$. It's integral from 1 to 2 is the same as every integral of a interval larger than $[1,2]$ and is equal to $\sum_{k=1}^n \frac{1}{n} \cdot \frac{1}{1+\frac{k}{n}}$. For the convergence I'm not sure how to deal with the sum, as I would normally start with finding the function it is pointwise converging to by calculating $\lim\limits_{n \to \infty} \varphi_n(x)$. – Tim Feb 03 '19 at 10:07
  • So far, so good. :-) Now, by how much can $\varphi_n$ and $f$ differ in the interval $[1+\frac{k-1}{n},1+\frac{k}{n})$?

    PS: Just noticed that it should be $[1+\frac{k-1}{n},1+\frac{k}{n})$ and not $[1+\frac{k-1}{n},1+\frac{k}{n}]$ in the definition of $\varphi_n$.

     – Mars Plastic Feb 03 '19 at 12:35
  • As $f=\frac{1}{x}$ is equal to the step function at $1+\frac{k}{n}$ and $\varphi_n$ is constant in $\left[1+\frac{k-1}{n},1+\frac{k}{n}\right)$ while $f$ is decreasing, the maximum difference between $\varphi_n$ and $f$ is at the "start" of the interval, $1+\frac{k-1}{n}$. Here, $f = \frac{1}{1+\frac{k-1}{n}}$ while $\varphi_n = \frac{1}{1+\frac{k}{n}}$. So the maximum difference is $f - \varphi_n = \frac{1}{1+\frac{k-1}{n}} - \frac{1}{1+\frac{k}{n}}$. Therefore in each interval it applies: $\lim\limits_{n \to \infty} \lVert \varphi_n - f \rVert_\infty = 0$, thus $\varphi_n \to f$ uniformly. – Tim Feb 03 '19 at 13:21
  • There you go. :-) But to be more precise, you should give a bound that does not depend on $k$. – Mars Plastic Feb 03 '19 at 13:23
  • Thank you very much for your help and your patience! – Tim Feb 03 '19 at 13:24
  • You're welcome! – Mars Plastic Feb 03 '19 at 13:29