Prove $ \sum\frac{1}{n} $ diverges by comparing with $\sum a_n$ where $a_n$ is the sequence $(\frac{1}{2}, \frac{1}{4}, \frac{1}{4}, \frac{1}{8}, \frac{1}{8}, \frac{1}{8}, \frac{1}{8},\frac{1}{16}, \frac{1}{16}, \frac{1}{16}, \frac{1}{16}, \frac{1}{16}, \frac{1}{16}, \frac{1}{16},\frac{1}{16}, \frac{1}{32}, \frac{1}{32}, ....)$
Another Proof that harmonic series diverges.
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calculus
sequences-and-series
analysis
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1In a [related thread](http://math.stackexchange.com/questions/255/why-does-the-series-frac-1-1-frac-12-frac-13-cdots-not-converge) you can find several proofs, including this one. – 2012-10-18
1 Answers
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This is one of the standard ways to prove that the harmonic series diverges.
HINT
- Find the general form of $a_n$.
- Prove that $\dfrac1n > a_n$
- Find $\displaystyle \sum_{n=1}^{n=2^m-1} a_n$
- From ($2$), we have $\displaystyle \sum_{n=1}^{n=2^m-1} \dfrac1n > \displaystyle \sum_{n=1}^{n=2^m-1} a_n$
- Conclude that $\displaystyle \sum_{n=1}^{n=\infty} \dfrac1n $ diverges by letting $m \to \infty$ in ($4$).
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0Can you give me a hint on how to find the general form on $a_n$ ? – 2012-10-18