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

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

• 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

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.
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.