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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    $\begingroup$ What is the question? $\endgroup$ Sep 23, 2016 at 18:28
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    $\begingroup$ The series is called the Harmonic series. It is notable for being divergent. $\endgroup$ Sep 23, 2016 at 18:28
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    $\begingroup$ I disagree to downvote a high school student. $\endgroup$
    – Jean Marie
    Sep 23, 2016 at 19:08
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    $\begingroup$ See Why does the series $\sum_{n=1}^\infty\frac1n$ not converge? (and perhaps also other questions linked there.) $\endgroup$ Sep 23, 2016 at 22:53
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    $\begingroup$ @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. $\endgroup$
    – Alex M.
    Sep 24, 2016 at 11:42

2 Answers 2


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.

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    $\begingroup$ +1. Good answer because it's expressed in an elementary jargon-free way that many people (including the OP) will probably understand. $\endgroup$
    – bubba
    Sep 24, 2016 at 11:29

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.

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    $\begingroup$ The 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. $\endgroup$
    – Mark Viola
    Sep 23, 2016 at 20:47
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    $\begingroup$ The Cauchy Condensation Test is a good test to show divergence here. It is equivalent to the demonstration in my answer. $\endgroup$
    – robjohn
    Sep 23, 2016 at 23:05
  • $\begingroup$ @robjohn Thanks. I knew the proof with the powers of two, but I didn’t know that it was the Cauchy Condensation Test. $\endgroup$ Sep 23, 2016 at 23:09

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