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Possible Duplicate:
Why does the series $\frac 1 1 + \frac 12 + \frac 13 + \cdots$ not converge?
Prove that the sequence converges

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?

Community
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phi
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    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
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    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
  • i have a bunch of exercises similar to this one, is there any technique for approaching them? – phi Jan 13 '13 at 21:00
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    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
  • http://math.stackexchange.com/questions/172504/

    http://math.stackexchange.com/questions/89799/

    http://math.stackexchange.com/questions/255/

    http://math.stackexchange.com/questions/56688/

    http://math.stackexchange.com/questions/55358/

    http://math.stackexchange.com/questions/215907/

     –  Jan 13 '13 at 21:01
  • I am voting to close as an exact duplicate. This question is covered in multiple places on the site, but I chose the question with the most easily understood answer. – JavaMan Jan 13 '13 at 21:01
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    Is the number of duplicates of this question bounded above? – David Mitra Jan 13 '13 at 21:01
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    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.

     –  Jan 13 '13 at 21:08

3 Answers3

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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.

Diego Silvera
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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.

Michael K
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Any sequence bounded above satisfies $X_n<n$ for large enough $n$, so it definitely cannot work.

A possibility is to write $$1+ \frac{1}{2}+...+ \frac{1}{n} \geq 1+ \int_1^n \frac{dx}{x}$$

Seirios
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