How to prove $\sum\limits_{k=2}^{n}\dfrac{1}{k}<\log(n)<\sum\limits_{k=1}^{n-1}\dfrac{1}{k}$

Question

How to prove $\sum\limits_{k=2}^{n}\dfrac{1}{k}<\log(n)<\sum\limits_{k=1}^{n-1}\dfrac{1}{k}$

It is clear if i consider the area under $f(x)=\dfrac{1}{x})$ from $1$ to $n$ end divide the interval $[1,n]$ into subintervals of length $1$, since $1/x$ is strictly decreasing LHS of the inequality takes the minimum value and RHS the maximum, so $\log(n)$ is always between them.

But I want to show that with Riemann-Sum, is this possible ?

I tried:

$(f(x):=\dfrac{1}{x})$

$$\log(n)={\displaystyle\int_{1}^{n}\dfrac{dx}{x}}=\lim\limits_{k\to\infty}\dfrac{n-1}{k}\sum\limits_{i=1}^{k}f\left(1+\dfrac{i(n-1)}{k}\right)=\lim\limits_{k\to\infty}\sum\limits_{i=1}^{k}\dfrac{n-1}{k+i(n-1)}$$

Now I'm stuck, how can i compare this with the sums above

Answer 1

Perhaps you're trying to get at the Riemann sums resulting from subdividing each interval $[k, k+1]$ into $m$ equal-length pieces? This gives $$ \frac{1}{k+1} = \sum_{i=1}^m \frac{1}{m} \cdot \frac{1}{k+1} < \sum_{i=1}^m \frac{1}{m} \cdot \frac{1}{k + \frac{i}{m}} < \sum_{i=1}^m \frac{1}{m} \cdot \frac{1}{k} = \frac{1}{k}, $$ so that $$ \sum_{k=1}^{n-1} \frac{1}{k+1} < \sum_{k=1}^{n-1} \sum_{i=1}^m \frac{1}{m} \cdot \frac{1}{k + \frac{i}{m}} = \frac{1}{m} \sum_{j=1}^{m(n-1)} \frac{1}{1 + \frac{j}{m}} < \sum_{k=1}^{n-1} \frac{1}{k}, $$ and $$ \frac{1}{m} \sum_{j=1}^{m(n-1)} \frac{1}{1 + \frac{j}{m}} = \sum_{j=1}^{m(n-1)} f(x_{j})\, \Delta x_{j} \approx \int_{1}^{n} \frac{1}{x}\, dx. $$

Answer 2

For $n>1$ $$\log(n)=\int_1^n\frac{1}{x}dx$$ thus $$\sum_{k=2}^n\frac{1}{k}\leq\log(n)\leq\sum_{k=2}^n\frac{1}{k-1}=\sum_{k=1}^{n-1}\frac{1}{k}.$$