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Prove $$ \sum\frac{1}{n} $$ diverges by comparing with $$\sum a_n$$ where $a_n$ is the sequence $$(\frac{1}{2}, \frac{1}{4}, \frac{1}{4}, \frac{1}{8}, \frac{1}{8}, \frac{1}{8}, \frac{1}{8},\frac{1}{16}, \frac{1}{16}, \frac{1}{16}, \frac{1}{16}, \frac{1}{16}, \frac{1}{16}, \frac{1}{16},\frac{1}{16}, \frac{1}{32}, \frac{1}{32}, ....)$$

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Where are you stuck? – Graphth Oct 17 '12 at 21:18
Well, It is obvious that $a_n < 1/n$. But how can I make this rigorous? – ILoveMath Oct 17 '12 at 21:20
I think this is the most frequently seen proof. (But others exist, and some of them are----or at least one of them is----just as simple and elementary.) – Michael Hardy Oct 17 '12 at 21:33
In a related thread you can find several proofs, including this one. – Martin Sleziak Oct 18 '12 at 10:14
up vote 3 down vote accepted

This is one of the standard ways to prove that the harmonic series diverges.


  1. Find the general form of $a_n$.
  2. Prove that $$\dfrac1n > a_n$$
  3. Find $$\displaystyle \sum_{n=1}^{n=2^m-1} a_n$$
  4. From ($2$), we have $$\displaystyle \sum_{n=1}^{n=2^m-1} \dfrac1n > \displaystyle \sum_{n=1}^{n=2^m-1} a_n$$
  5. Conclude that $$\displaystyle \sum_{n=1}^{n=\infty} \dfrac1n $$ diverges by letting $m \to \infty$ in ($4$).
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Can you give me a hint on how to find the general form on $a_n$ ? – ILoveMath Oct 18 '12 at 3:12

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