Timeline for Awfully sophisticated proof for simple facts
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
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| Nov 6 at 9:11 | comment | added | FShrike | @Smiley1000 can’t actually dominate these guys in the obvious way given the geometric series converges. For the harmonic example the gaps between each term are $1$, so you can just use the series. But in this case, I might put $1$ on $[0,1]$, $1/2$ on $[1,2]$, $1/4$ on $[2,4]$, $1/8$ on $[4,8]$, $1/16$ on $[8,16]$… and because of the increasingly large gaps, the integral of this function comes to $1+1/2+1/2+1/2+…=\infty$ just like in the classical proof the harmonic series diverges | |
| Jul 31 at 10:02 | comment | added | Smiley1000 | @user56097 Can you explain why the argument doesn't work for the subsequence? | |
| Mar 23, 2019 at 14:42 | review | Suggested edits | |||
| Mar 23, 2019 at 15:25 | |||||
| Jul 18, 2014 at 21:17 | comment | added | user56097 | I love this proof, thank you very much. I explained it to some students, and challenged them saying "and since the contradiction appears along any subsequence (for example the powers of 2), the series of the $2^{-n}$ also diverges". It took them a minute's thought to find the error. (One can play a bit further with the proof to show that if $\sum_n a_n$ is a divergent series with positive terms, then $\sum_n \frac{a_n}{a_1+\dots+a_n}$ also diverges. Applying it with $a_n=1$ gives the divergence of the harmonic series. Then, taking $a_n=1/n$ gives the divergence of $\sum_n \frac{1}{n H_n}$.) | |
| Feb 24, 2013 at 23:29 | comment | added | Jesse Madnick | And here I was thinking the standard proof was just the Integral Test for series convergence. | |
| Nov 28, 2011 at 9:42 | comment | added | John Jiang | I remember someone had an article giving 20 different proofs of this fact. | |
| Jul 8, 2011 at 22:09 | comment | added | The Mathemagician | @Kevin Actually,Gerald Itzkowitz once showed me that proof in his integration theory course. | |
| Jul 8, 2011 at 21:19 | history | edited | Suvrit | CC BY-SA 3.0 |
fixed misplaced comma
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| Jun 19, 2011 at 22:26 | comment | added | Kevin O'Bryant | The standard proof is $\sum_{n=2^i+1}^{2^{i+1}} n^{-1} \geq \sum_{n=2^i+1}^{2^{i+1}} 2^{-(i+1)} = \frac 12$. | |
| Jun 5, 2011 at 16:25 | history | edited | C.S. | CC BY-SA 3.0 |
added 2 characters in body
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| Dec 9, 2010 at 13:13 | comment | added | Mariano Suárez-Álvarez | @Harry, no it isn't: this depends on knowing the dominated convergence theorem (which very few people prove for the Riemann integral, so usually has to wait until you are studying measure theory) The divergence of the harmonic series follows from the integral comparison thorem, for example, a much more elementary proof! | |
| Dec 9, 2010 at 6:03 | comment | added | roy smith | Good point, but I guess the standard proof is that the series dominates the sum of the functions (1/2)^(n+1).1([2^n, 2^(n+1)]. Is that equivalent? | |
| Dec 9, 2010 at 4:58 | comment | added | Harry Gindi | Isn't this the standard proof? | |
| Nov 4, 2010 at 3:17 | comment | added | Mariano Suárez-Álvarez | I love this one. | |
| Nov 3, 2010 at 22:19 | history | answered | Terry Tao | CC BY-SA 2.5 |