Why compare $s_n:=\sum_{k=1}^n\frac{1}{k}$ to $\int_{1}^{n+1}\frac{1}{x}dx$, instead of $\int_1^{n}\frac1{x}dx$, when proving divergence of $s_n$?

Question

I've seen an example in which the limit of the sequence $s_n = \sum_{k=1}^n \frac{1}{k}$ is proved to be divergent because:

$$s_n \geq \int_{1}^{n+1}\frac{1}{x} dx$$

and $\log(n+1)=+\infty$ as $n\to +\infty$. I'm confused about the following: Why $n+1$ instead of $n$? I've been trying to think but found no (convincent) reason for it, I guess it would work with $n$ too.

I guess I got confused because in the previous example, he proved that $s_n=\sum_{k=1}^{n}\frac{1}{k^2}$ is convergent because:

$$s_n\leq 1+\int_1^{n}\frac{1}{x^2}dx$$

The number of unit lengths between $1$ and $n$ is only $n-1$, so you would not compare it with $n$ terms. — Jair Taylor, Jun 08 '19 at 00:29

score 12 · Answer 1 · answered Jun 08 '19 at 00:26

12

$\int_1^{n+1}\frac 1 x dx=\sum_{k=1}^{n} \int_k^{k+1} \frac 1 x dx \leq \sum_{k=1}^{n} \frac 1 k$ because $\frac 1 x \leq \frac 1k$ when $x$ is between $k$ and $k+1$.

answered Jun 08 '19 at 00:26

Kavi Rama Murthy

311,013

score 1 · Answer 2 · answered Jun 08 '19 at 12:04

Actually it is OK to use $\int_1^n{1\over x}\,dx=\ln n$ for the comparison, it just requires the extra step of noting that, since the integrand $1\over x$ is non-negative for $x\gt0$, we have $\int_1^{n+1}{1\over x}\,dx\ge\int_1^n{1\over x}\,dx$. Alternatively, one can use the non-negativity of the terms in the series to say $\sum_{k=1}^n{1\over k}\ge\sum_{k=1}^{n-1}\ge\int_1^n{1\over x}\,dx$.

The point is, the key comparison interprets each each term in the series as the area beneath a single step in a step function, each step being of width $1$, so the natural comparison for a sum of $n$ terms is an integral over an interval of length $n$.

Why compare $s_n:=\sum_{k=1}^n\frac{1}{k}$ to $\int_{1}^{n+1}\frac{1}{x}dx$, instead of $\int_1^{n}\frac1{x}dx$, when proving divergence of $s_n$?

2 Answers2