1

I understand that the harmonic series diverges because of the comparison test: $$1+\frac12+\frac13+\frac14+\frac15+\frac16+\frac17+\frac18+\cdots\\\ge1+\frac12+\frac14+\frac14+\frac18+\frac18+\frac18+\frac18+\cdots$$However, I cannot intuitively understand why this is the case if $$\lim_{n \to \infty}\left(\frac{1}{n}\right)=0$$ Previosly, I thought an easy way of checking whether a series is convergent is if the above condition is true. However, the harmonic series has shown to me that this is not always the same the case—sometimes a series can diverge even if the terms approach $0$. Why does the harmonic series diverge but series such as $1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots$ do not?

Joe
  • 19,636
  • Recall that $$\sum_{n = 1}^{+\infty} a_n = s \neq \pm \infty \Rightarrow \lim_{n \to +\infty} a_n = 0.$$ The vice versa is in general not true. – the_candyman Dec 21 '19 at 19:02
  • 2
    The harmonic series diverges, because it diverges, the series $\sum_n 2^{-n}$ doesn't diverge because it converges. Is there much else to be said about this? – Angina Seng Dec 21 '19 at 19:02
  • 1
    Essentially it is how fast the terms $\to 0$, not just that they do. – herb steinberg Dec 21 '19 at 19:04
  • $1+\frac12+\frac14+\frac14+\frac18+\frac18+\frac18+\frac18+\cdots$ diverges (add up an infinite number of $\frac12$s). But the individual terms decrease towards $0$ – Henry Dec 21 '19 at 19:05
  • @ Lord Shark the Unknown But I'm interested as to why the harmonic series diverges. I accept that it is true because of the comparison test, but I would like this to align with my other intuitions, so I can have a better handle on infinite series. – Joe Dec 21 '19 at 19:28
  • 1
    Do you think $1+1+ 1 + 1 +\cdots $ is finite? After all, isn't this much the same as

    $$1 + \frac{1}{2} + \frac{1}{2}+ \frac{1}{3}+ \frac{1}{3}+ \frac{1}{3} + \frac{1}{4}+ \frac{1}{4}+ \frac{1}{4}+ \frac{1}{4} +\cdots$$

    In this latter series, the terms are positive and $\to 0,$ although at a slower rate.

     – zhw. Dec 22 '19 at 02:44

2 Answers2

1

How about this example, $\sqrt n\to\infty$ as $n\to\infty$, as I hope you'd agree? Let $a_1=1$, $a_2=\sqrt2-1$, $a_3=\sqrt3-\sqrt2,\ldots,a_n=\sqrt n-\sqrt{n-1},\ldots$. Then $a_1+\cdots+a_n=\sqrt n$, and so the series $\sum_{n=1}^\infty a_n$ is divergent. But you can prove that $a_n$ is approximately $\frac1{2\sqrt n}$ and that $a_n\to0$ as $n\to\infty$. So there's another series whose terms tend to zero, but is divergent. I hope this isn't too painful.

Angina Seng
  • 158,341
1

The implication only goes one way: if $\sum_n a_n<\infty$, then $a_n\to 0$, but not the other way around. As your examples show, if $a_n\to 0$ then both convergence and divergence of the series are possible.

Instead, it has to do with the speed at which $a_n$ converges to $0$. Exponential convergence (like the $1+\tfrac12+\tfrac14+\cdots$ series you wrote) is extremely fast, so the series converges. Polynomial convergence (like $\sum_n 1/n^2,\sum_n 1/n^3$, etc) is a lot slower than exponential, but fast enough for the sum to converge when the power is greater than $1$. But it is too slow when the power is $1$ or smaller.

The intuition behind this is that when you sum a polynomial expression, its degree increases by $1$. That's why the cutoff is at degree $-1$, since when you sum it the cutoff translates to degree $0$ polynomials, which on the one side (slightly negative) stays bounded, and on the other side (slightly positive) tends to $\infty$.

pre-kidney
  • 30,223