the difference between discrete sum and integral

Question

hi i have as question

we know that that $\frac1x$ is a strictly decreasing function on $[1,\infty[$

let n a natural number greater than 0

let $u_n=\sum_{i=n}^{2n} \frac1i$

and $F(n)=\int_{n}^{2n} \frac1x dx$

i want to prove that $\frac1{2n}\le u_n-F(n)\le \frac1n$

i have tried to transform the integral into a sum but i couldn't figure it out

please help me with this question

Have you tried approximating the integral using rectangles:
$$F(n)=\int_n^{2n}\frac{1}{x}dx\approx \sum_{i=1}^m \frac{1}{t_i}(x_{i+1}-x_i)$$

where $x_0=n\leq x_1\leq x_2 \leq \dots \leq x_m=2n$, and $t_i\in [x_{i+1}-x_i]$. — Weierstraß Ramirez, May 31 '23 at 10:11
@WeierstraßRamirez we have to find an inequality i had not get the idea to approxiamte the integral — user912835, May 31 '23 at 10:15

score 1 · Accepted Answer · answered May 31 '23 at 09:33

1

Hint : Decompose $F(n)$ into a sum of integrals between two consecutive integers.

answered May 31 '23 at 09:33

nicomezi

do you mean $F(n)=\sum_{i=n}^{2n-1}\int_{i}^{i+1} \frac1x dx$ – ana nadi lwa3r May 31 '23 at 09:41
Yes, then you have obvious upper and lower bound of each integrals since $1/x$ is monotoneous. – nicomezi May 31 '23 at 10:25

1 Answers1