Prove $\int^{n+1}_1\frac{1}{x} dx \leq 1 + 1/2 + \cdots + 1/n$.

Question

Source: https://math.mit.edu/~choiks/Pset%202%20solutions.pdf

I don't understand the third line $\int^{n+1}_1\frac{1}{x}dx \leq 1 + 1/2 + \cdots + 1/n$.

I think it should rather be $\int^{n+1}_1\frac{1}{x}dx \geq 1 + 1/2 + \cdots + 1/n$ because

$\int^{n}_1\frac{1}{x} < \int^{n+1}_1\frac{1}{x}$ and $1 + 1/2 + \cdots + 1/n$ is just addition of $f(x) = 1/x$ at certain values (integers). Therefore:

$1 + 1/2 + \cdots + 1/n < \int^{n}_1\frac{1}{x} < \int^{n+1}_1\frac{1}{x}$.

Thank you!

Answer 1

Consider this example:

$A_i =$ the area under $\frac 1x$ from $x=1$ to $x=3$. $A_a =$ the area of the width-$1$ rectangle, with its base on the $x$-axis, and its upper-left corner on the point $(1,1)$, plus the area of the width-$1$ rectangle, with its base on the $x$-axis, and its upper-left corner on the point $(2,\frac 12)$.

Certainly (and this may be clearer if you draw the rectangles on the curve) we have $A_i < A_a$. That is, $\int_0^{2+1} \frac 1x dx < 1 + \frac 12$. Loosely speaking, this is because the two rectangles together a) cover up the entire area enclosed by $y=\frac 1x,y=0,x=1,x=3$, and then have some 'area left over.' Indeed, these estimations with rectangles are left Riemann sums, which are overestimations if you use them to estimate functions which are monotonic decreasing on an interval. The link provided has better (visual!) examples.

In this case, this generalizes to $\int_1^{n+1} \frac 1x dx < \sum_{k <n+1} \frac 1k$ for $n \geq 1$, which is what you were looking for.

Answer 2

$$\int_1^{n+1}\dfrac{1}{x}\mathrm dx=\sum_{k=1}^n\int_k^{k+1}\dfrac{1}{x}\mathrm dx$$ $$<\sum_{k=1}^n\int_k^{k+1}\dfrac{1}{k}\mathrm dx$$ $$=\sum_{k=1}^n\dfrac{1}{k}$$ Similarly, you use $$\sum_{k=1}^n\int_k^{k+1}\dfrac{1}{x}\mathrm dx>\sum_{k=1}^n\int_k^{k+1}\dfrac{1}{k+1}\mathrm dx$$ to get the other bound.

Answer 3

There are simple diagrams that compare sums and integrals.... Here what I call $f(x)$ is real valued and continuous, also positive. Things change (order of the inequalities) depending on $f$ increasing or decreasing. I guess $a < b$ are fixed integers.

if we have $f(x) > 0$ and $f'(x) > 0,$ then $$ \int_{a-1}^{b} \; f(x) \; dx \; < \; \sum_{j=a}^b \; f(j) \; < \; \int_{a}^{b+1} \; f(x) \; dx $$

if we have $f(x) > 0$ but $f'(x) < 0,$ then $$ \int_a^{b+1} \; f(x) \; dx \; < \; \sum_{j=a}^b \; f(j) \; < \; \int_{a-1}^b \; f(x) \; dx $$ We use this second one, as we are going to integrate $1/x$ we take $a=2$ along with $b=n$ $$ \int_2^{n+1} \; \frac{1}{x} \; dx \; < \; \sum_{j=2}^n \; \frac{1}{j} \; < \; \int_{1}^n \; \frac{1}{x} \; dx $$ $$ 1+ \int_2^{n+1} \; \frac{1}{x} \; dx \; < \; 1+ \sum_{j=2}^n \; \frac{1}{j} \; < \; 1+ \int_{1}^n \; \frac{1}{x} \; dx $$ $$ 1+ \log (n+1) - \log 2 \; < \; \sum_{j=1}^n \; \frac{1}{j} \; < \; 1+ \log n $$