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?

marked as duplicate by JavaMan, David Mitra, user17762, leonbloy, Ayman HouriehJan 13 '13 at 21:05

• 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 $1$. – Brian M. Scott Jan 13 '13 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. – Sangchul Lee Jan 13 '13 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 '13 at 21:01
• Is the number of duplicates of this question bounded above? – David Mitra Jan 13 '13 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 '13 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 $$1+ \frac{1}{2}+...+ \frac{1}{n} \geq 1+ \int_1^n \frac{dx}{x}$$