The harmonic series isn't bounded

[Jamin2112]

Homework Statement

My mind is blown. You'd think there would be some number which 1/1 + 1/3 + 1/4 + ... stays below, but I guess there isn't. However, before I believe this, I need one part of my book's proof clarified.

Homework Equations

Theorem I. Suppose that u_n ≥ 0 for every n. Then the series ∑u_n is convergent if and only if the sequence {s_n} of partial sums is bounded.

The Attempt at a Solution

I'm just following along the proof in the book.

s_n = 1 + 1/2 + ... + 1/n.

s_2n = s_n + 1/(n+1) + 1/(n+2) + ... + 1/2n > s_n + 1/2,

since

1/(n+1) + 1/(n+2) + ... + 1/2n > n * (1/2n) = 1/2.

^ That's the part I don't understand. Please explain why that is true.

 

[Office_Shredder]

$$\frac{1}{n+1}\geq\frac{1}{2n}$$. Also $$\frac{1}{n+2}\geq\frac{1}{2n}$$, $$\frac{1}{n+3}\geq\frac{1}{2n}$$,..., $$\frac{1}{2n}\geq\frac{1}{2n}$$.

So adding up each term between 1/(n+1) and 1/2n is adding up n terms, each of which is at least 1/2n. The sum must then be at least 1/2n+1/2n+1/2n+...+1/2n=n*(1/2n)=1/2

 

[fzero]

1/(n+1) + 1/(n+2) + ... + 1/2n > n * (1/2n) = 1/2.

^ That's the part I don't understand. Please explain why that is true.

Each term in the sequence on the LHS is greater than the next, so each term is greater than 1/(2n):

$$\frac{1}{n+1} > \frac{1}{n+2}, ~ \frac{1}{n+2} > \frac{1}{n+3}, \ldots \frac{1}{2n-1} > \frac{1}{2n}.$$

It follows that the sum of all terms is greater than n(1/(2n)).

 

[Jamin2112]

$$\frac{1}{n+1}\geq\frac{1}{2n}$$. Also $$\frac{1}{n+2}\geq\frac{1}{2n}$$, $$\frac{1}{n+3}\geq\frac{1}{2n}$$,..., $$\frac{1}{2n}\geq\frac{1}{2n}$$.

So adding up each term between 1/(n+1) and 1/2n is adding up n terms, each of which is at least 1/2n. The sum must then be at least 1/2n+1/2n+1/2n+...+1/2n=n*(1/2n)=1/2