# How can $\sum_{n=1}^\infty \frac{1}{n}$ be simplified [duplicate]

This question already has an answer here:

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:

$$\sum_{n=1}^\infty \frac{1}{n}$$

## marked as duplicate by 6005, Joey Zou, quid♦, Martin Sleziak, Jack's wasted lifeSep 24 '16 at 19:33

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• What is the question? – Robert Israel Sep 23 '16 at 18:28
• The series is called the Harmonic series. It is notable for being divergent. – Matthew Leingang Sep 23 '16 at 18:28
• I disagree to downvote a high school student. – Jean Marie Sep 23 '16 at 19:08
• See 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
• @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

## 2 Answers

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.

• +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

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.

• 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. – Mark Viola Sep 23 '16 at 20:47
• The 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
• @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