So I'm doing year 9/10 now and I've just been working with sigmas $\Sigma$.
I found across a question which I found quite tricky.
Is there a way to write down the answer to this question or make it easier:
$$\sum_{n=1}^\infty \frac{1}{n}$$
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5What is the question? – Robert Israel Sep 23 '16 at 18:28
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6The series is called the Harmonic series. It is notable for being divergent. – Matthew Leingang Sep 23 '16 at 18:28
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there are many articles about the harmonic series in the internet – Dr. Sonnhard Graubner Sep 23 '16 at 18:37
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1It's not clear what exactly you're asking – Ben Grossmann Sep 23 '16 at 18:47
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23I disagree to downvote a high school student. – Jean Marie Sep 23 '16 at 19:08
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1@Omnomnomnom, I suspect he assumes there is some sort of closed form for this expression, like for many other series, for example sum of the first n integers. Closed form here, a formular can just type in a calculator in less than 10 seconds to solve a problem, which would take ten times longer normally. – Imago Sep 23 '16 at 19:11
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2It can be simplified as $\infty$. – Sep 23 '16 at 19:47
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1@YvesDaoust One should always be reluctant to just simplify a sum to infinity. For example, $\frac{\sum_{n=1}^\infty \frac 2 n}{\sum_{n=1}^\infty \frac 1 n} \rightarrow \frac \infty \infty \rightarrow ?$. – Axoren Sep 23 '16 at 19:51
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7See Why does the series $\sum_{n=1}^\infty\frac1n$ not converge? (and perhaps also other questions linked there.) – Martin Sleziak Sep 23 '16 at 22:53
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@PichiWuana The first edit after the question is put on hold puts question into reopen review queue. (For more details see here.) In this case, the result was to leave it closed. It is advisable that the first edit after putting on hold is substantial enough that it addresses the issues why the post has been put on hold. – Martin Sleziak Sep 24 '16 at 06:19
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4@Axoren: it is a well-known fact that $\frac\infty\infty=2$. – Sep 24 '16 at 07:48
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@MartinSleziak oh ok didn't know that – Pichi Wuana Sep 24 '16 at 09:39
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1I don't see anything at all wrong with this question. The answer is fairly well known to people who have taken a few maths courses, but not to 10-th graders. – bubba Sep 24 '16 at 11:36
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10@YvesDaoust: You should not joke like this: less experienced people might take you seriously just because you have a high MSE reputation. Or at least use smileys. – Alex M. Sep 24 '16 at 11:42
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5@JeanMarie Any high school student? Whichever question they ask? This seems utterly unreasonable... – Did Sep 24 '16 at 16:01
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1@AlexM. Especially since $\frac\infty\infty=42$. – Did Sep 24 '16 at 16:02
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3I reopened, then voted to close as duplicate, just so we can get the close reason right on the cleaned-up question. – Caleb Stanford Sep 24 '16 at 17:26
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2Since I can't add an answer here, I added one to the duplicate question: http://math.stackexchange.com/questions/255/why-does-the-series-sum-n-1-infty-frac1n-not-converge?noredirect=1&lq=1. It's very similar to some other answers, but with special emphasis on making it understandable to a 9th grade student. – bubba Sep 25 '16 at 02:25
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@AlexM.: you are right, I forgot to smile. – Sep 25 '16 at 19:40
2 Answers
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Consider breaking the sum, after the first term, into chunks with $2^{n-1}$ terms ending in $\frac1{2^n}$ $$ 1+\overbrace{\ \ \ \ \ \frac12\ \ \ \ \ }^{\ge1/2}+\overbrace{\ \frac13+\frac14\ }^{\ge1/2}+\overbrace{\frac15+\frac16+\frac17+\frac18}^{\ge1/2}+\overbrace{\frac19+\frac1{10}+\cdots+\frac1{16}}^{\ge1/2}+ $$ You can see that we can add as many chunks as we wish that are at least as large as $\frac12$. By continuing in this fashion, the sum can be made as large as we want.
robjohn
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2+1. Good answer because it's expressed in an elementary jargon-free way that many people (including the OP) will probably understand. – bubba Sep 24 '16 at 11:29
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You can use the fact that the limiting difference between the natural logarithm and the Harmonic Series is the Euler-Mascheroni Constant to estimate the $nth$ term of the series you are using. The only problem is that the Harmonic Series diverges, which you can prove using the integral test or various other tests.
AlgorithmsX
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2The ratio test is inconclusive in proving divergence of the Harmonic series. And "other tests" fail also. There are several ways aside from the integral test, that do demonstrate divergence. – Mark Viola Sep 23 '16 at 20:47
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2The Cauchy Condensation Test is a good test to show divergence here. It is equivalent to the demonstration in my answer. – robjohn Sep 23 '16 at 23:05
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@robjohn Thanks. I knew the proof with the powers of two, but I didn’t know that it was the Cauchy Condensation Test. – AlgorithmsX Sep 23 '16 at 23:09