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I can't understand intuitively why the series

$$ \sum^{\infty}_{k=1} \frac{1}{k} $$

diverges while its counterpart with only the alternating signs does the opposite (converges)

$$\sum^{\infty}_{k=1} \frac{(-1)^k}{k} $$

Can someone explain?

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    Your calculus text should have a section on "conditional convergence" of series where this sort of thing is discussed. – GEdgar Apr 02 '23 at 00:24
  • $$\sum_{k=1}^n \frac{1}{k} > \ln(n) \implies \lim_{n\to\infty} \sum_{k=1}^n \frac{1}{k} > \lim_{n\to\infty} \ln(n).$$ $$\ln(2) = \int_0^1 \frac{dx}{x+1} = \int_0^1 \left[ ~1 - x + x^2 - x^3 + \cdots - \cdots ~\right] ~dx.$$ The RHS expression above integrates to $$\lim_{n\to\infty} \sum_{k=1}^n \frac{(-1)^{k+1}}{k}.$$ – user2661923 Apr 02 '23 at 01:12
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    You seem to be asking two unrelated questions which should be the subject of two separate posts, nsamely, why does the harmonic series diverge, and why does the alternating harmonic series converge. If you want to know why those two series behave differently, you would have to tell us why you think they should be the same, so we can explain what is wrong with your thinking. You have not bothered to tell us why you think introducing all those minus signs isn't going to change anything. – bof Apr 02 '23 at 03:19
  • Does this answer your question? Change of signs in harmonic series – Anne Bauval Apr 02 '23 at 07:01

1 Answers1

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This is a classic result. I am sure this is answered somewhere else.

My math professor's proof, I think is very intuitive:

$ \frac{1}{1} + \frac{1}{2} + \big(\frac{1}{3} + \frac{1}{4} \big)+ \big(\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8}\big) + ... > \frac{1}{1} + \frac{1}{2} + \big(\frac{1}{4} + \frac{1}{4} \big)+ \big(\frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8}\big) + ...$

$ \Rightarrow \frac{1}{1} + \frac{1}{2} + \big(\frac{1}{3} + \frac{1}{4} \big)+ \big(\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8}\big) + ... > \frac{1}{1} + \frac{1}{2} + \big(\frac{1}{2} \big)+ \big(\frac{1}{2} \big) + ...$

The right hand side clearly diverges, and thus the original sum diverges. You are adding smaller and smaller pieces, but the sum of an increasing number of those pieces always remains finite (sum of $2^n$ terms will always give you 1/2).

The alternating counterpart can be shown to be bounded using a similar argument: $ \frac{1}{1} - \big(\frac{1}{2} - \frac{1}{3} \big) - \big(\frac{1}{4} - \frac{1}{5}\big) - \big(\frac{1}{6} - \frac{1}{7}\big) ... < \frac{1}{1} - \big(\frac{1}{2} - \frac{1}{2} \big) - \big(\frac{1}{4} - \frac{1}{4}\big) - \big(\frac{1}{6} - \frac{1}{6}\big) + ... = 1$

Thus, the alternating series is bound by 1. Here you are subtracting something and then immedietely adding something slightly smaller (the way its grouped above). The net subtraction amount keeps reducing at a rate that is equal to $O(1/n^2)$. The actual value of the alternating series can be obtained from the taylor series of ln(1+x) at x=1.

Hope this helps.

solving_math_equations
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  • We are not supposed to answer duplicates (I proposed one but there are tons), and to such "no-effort" questions. – Anne Bauval Apr 02 '23 at 07:01