Famous harmonic series — solutions

Solutions to previous blog questions on harmonic series are presented below:

Basic Reference: Popular Problems and Puzzles in Mathematics by Asok Kumar Mallik, IISc Press, Foundation Books; Amazon India link:

https://www.amazon.in/Popular-Problems-Puzzles-Mathematics-Mallik/dp/938299386X/ref=sr_1_2?ie=UTF8&qid=1505622919&sr=8-2&keywords=popular+problems+and+puzzles+in+mathematics

Detailed Reference:

Mallik, A. K. 2007: “Curious consequences of simple sequences,” Resonance, (January), 23-37.

Personal opinion only: Resonance is one of the best Indian magazines/journals for elementary/higher math and physics. It behooves you to subscribe to it. It will help in RMO, INMO and Madhava Mathematics Competition of India.

http://www.ias.ac.in/Journals/Resonance_–_Journal_of_Science_Education/

Solutions:

1. The thirteenth century French polymath Nicolas-Oresme proved that the harmonic series :$1+ \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \ldots$ does not converge. Prove this result.

Solution 1:

Nicolas Oreme had provided a simple proof as it involves mere grouping of terms, noticing patterns and making comparisons:

$\lim_{n \rightarrow \infty}H_{n}=H_{\infty}=1+\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \ldots$

$> 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{4} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \ldots$

$> 1+ \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \ldots$

Therefore, $H_{\infty}$ diverges as we go on adding one half indefinitely. Here is another way to prove this:

Consider $\lim_{n \rightarrow \infty}H_{n}=H_{\infty}=1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \ldots$

By multiplying and dividing both sides by 2 and then by regrouping the terms, we get:

$H_{\infty}=\frac{2}{2} + \frac{2}{4} + \frac{2}{6} + \frac{2}{8} + \ldots$

$H_{\infty}= \frac{1+1}{2} + \frac{1+1}{4} + \frac{1+1}{6} + \frac{1+1}{8} + \ldots$

$H_{\infty}= (\frac{1}{2} + \frac{1}{2}) + (\frac{1}{4} + \frac{1}{4}) + (\frac{1}{6} + \frac{1}{6}) + (\frac{1}{8} + \frac{1}{8}) + \ldots$

$H_{\infty}<1+ \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \ldots$

leading to a contradiction that $H_{\infty} The contradiction arose because only finite numbers remain unaltered when multiplied and divided by 2. So, $H_{\infty}$ is not a finite number, that is, it diverges.

2. Prove that $\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \ldots$ does not converge.

Solution 2:

$H_{E}=\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \frac{1}{8} + \ldots=\frac{1}{2}(1+\frac{1}{2} + \frac{1}{3} + \frac{1}{4} +\ldots)=\frac{1}{2}H_{n}$.

Since $H_{\infty}$ diverges, so does $H_{E}$.

3. Prove that $1+ \frac{1}{3} + \frac{1}{5} + \frac{1}{7} + \ldots$ does not converge.

Solution 3:

$H_{O}=\frac{1}{1}+ \frac{1}{3} +\frac{1}{5}+ \frac{1}{7}+ \ldots$ diverges as each term in this series is greater than the corresponding term of $H_{E}$, which we have just sent to diverge.

Cheers,

Nalin Pithwa.

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