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This question came to me from one of my calculus students today: Other than using the integral test $$\int_1^\infty \frac{dx}{x} \to \infty,$$ what are some other ways that we can prove the Harmonic series $\sum_{n=1}^\infty \frac{1}{n}$ diverges?

I'm sure there are plenty of methods out there; any method where a typical student in Calculus could understand would be great.

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  • $\begingroup$ You may assign this as a homework for him to discover a proof on his own; maybe you thus discover another Milnor :) $\endgroup$
    – Yes
    Oct 28, 2015 at 4:24
  • $\begingroup$ yea probably something I can let him on his own a little bit. $\endgroup$ Oct 28, 2015 at 4:34
  • $\begingroup$ A good example for introducing the Cauchy Condensation test. $\endgroup$ Oct 28, 2015 at 19:31

4 Answers 4

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Try applying the Comparison test to the harmonic series, specifically to this series

$ 1 + 1/2 + 1/4 + 1/4 + 1/8 + 1/8 + 1/8 + 1/8 + 1/16 + ...$

= $1 + 1/2 + (1/4+1/4) + (1/8+1/8+1/8+1/8) + ...$

= $1+ 1/2 + 1/2 + 1/2 + ... = \infty$

Since each term is larger than the term in the above series, the harmonic series diverges as well.

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Suppose

$$S=1+{1\over2}+{1\over3}+{1\over4}+\cdots$$

is convergent. Since all the terms are non-negative, it must be absolutely convergent. But absolutely convergent sequences can be fiddled with at will, allowing us to conclude

$$\begin{align} 2S&=2+1+{2\over3}+{1\over2}+\cdots\\ &=S+\left(2+{2\over3}+{2\over5}+\cdots\right)\\ &\gt S+\left(1+{1\over2}+{1\over3}+\cdots\right)\\ &=2S \end{align}$$

which is absurd: No number is greater than itself.

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The "easiest" way in my opinion is to remark that

$$\sum_{k=n+1}^{2n} \frac{1}{k} \geq n\frac{1}{2n} = \frac{1}{2}$$

So if we cut the sum between $1$ and $2^n$ by powers of two, we have

$$\sum_{k=1}^{2^n} \frac{1}{n} = \sum_{k=1}^{n} \sum_{i=2^{k-1}+1}^{2^k} \frac{1}{i} \geq \sum_{k=1}^n \frac{1}{2} = \frac{n}{2}$$

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  • $\begingroup$ I saw this in a book as 1+1/2+1/3+1/4+...>1+1/2+(1/4+1/4)+(1/8+1/8+1/8+1/8)+...=1=1/2+1/2+.... $\endgroup$ Oct 28, 2015 at 19:26
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Consider the generating function $$ f(x) = \sum_{k=1}^\infty \frac{x^n}{n} $$ and differentiate in the radius of convergence to get $$ f'(x) = \frac{d}{dx} \sum_{k=1}^\infty \frac{x^n}{n} = \sum_{k=1}^\infty x^{n-1} = \frac{1}{1-x}, $$ which only converges for $|x| < 1$, so $f(x)$ will only be defined in the same interval, and the original series is $f(1)$, which diverges.

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