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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0Where are you stuck? – 2012-10-17
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0Well, It is obvious that $a_n < 1/n$. But how can I make this rigorous? – 2012-10-17
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2I 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.) – 2012-10-17
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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