# Showing that $X_n = 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}$ is not bounded above. [duplicate]

I have to show that $X_n$ is not bounded above,

$$0<1\le1$$ $$0<\frac{1}{2}<1$$ $$\vdots$$ $$0<\frac{1}{n}<1$$

Adding up the inequalities we get $0<X_n<n,\ and\ n\to\infty$ so $X_n$ is not bounded above. Is this any good?

• I’m afraid not: the same reasoning would lead you to the conclusion that $$\frac12+\frac14+\frac18+\frac1{16}+\ldots$$ was unbounded, but in fact it’s equal to $1$. Jan 13, 2013 at 20:57
• If your argument were write, then we can also conclude that $$Y_n = 1 + \frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{2^n}$$ is not bounded above since $0 < X_n < n$ for any $n \geq 2$. I hope that this example would illuminate which error you made. Jan 13, 2013 at 20:58
• Let's see, $1 < 10$, $1.1 < 100$, $1.11 < 1000$, $1.111 < 10000$, and so on, so I guess $1.11111...$ is infinite. Or maybe not...
– KCd
Jan 13, 2013 at 21:01
• Is the number of duplicates of this question bounded above? Jan 13, 2013 at 21:01
• To show that a series diverges, you need to bound it below by a divergent series. Similarly, to show that a series converges, you need to bound it above a convergent series.
– user17762
Jan 13, 2013 at 21:08

Observe this $$X_1 = 1$$ $$X_2 = 1 + \frac{1}{2}$$ $$X_4 = X_2 + \frac{1}{3} + \frac{1}{4} \geq 1 + \frac{1}{2} + \frac{2}{4} = 1 + \frac{1}{2} + \frac{1}{2}$$ $$X_8 = X_4 + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} \geq 1 + \frac{1}{2} + \frac{1}{2} + \frac{4}{8} = 1 + \frac{1}{2} + \frac{1}{2} + \frac{1}{2}$$ and so on.
Try adding together all the numbers x $1/2^n\le x < 1/2^{n+1}$ for $n\in\mathbb N$. Notice that there are infinite number of these partitions, so if you can bound the value of the sum of the numbers in each partition by some positive constant, then the sum of all the numbers must be infinite.
Any sequence bounded above satisfies $X_n<n$ for large enough $n$, so it definitely cannot work.
A possibility is to write $$1+ \frac{1}{2}+...+ \frac{1}{n} \geq 1+ \int_1^n \frac{dx}{x}$$