A Proof(??) of harmonic series being divergent

1+12+13+14+...=1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + ... =

(1+12+14+...12n)(1+13+19+...13n)...(1+1p+1p2+...1pn)(1 + \frac{1}{2} + \frac{1}{4} + ... \frac{1}{2^n})(1 + \frac{1}{3} + \frac{1}{9} + ... \frac{1}{3^n}) ... (1 + \frac{1}{p} + \frac{1}{p^2} + ... \frac{1}{p^n})

Where p'p' is a prime number.

As each of the sums in the brackets are >1, the product of all of these terms is divergent (as there are infinitely many primes and the product of infinite terms all >1 can't be finite*). Therefore, the harmonic series is divergent.

I decided to write this after reading other proofs of this. It's most likely already been done more rigorously centuries ago so if you know who first came up with a similar proof please post a link in the comment section. And also if anyone could actually prove point *. Sorry if the title ('proof') was misleading.

Note by Nucky Korprasertsri
5 years, 6 months ago

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You can also group these terms as such : 1 + (1/2) + (1/3+1/4) + ( 1/5+1/6+1/7)+( 1/8+1/9+1/10+1/11+1/12) and so forth... => 1 + 1/2+ 7/12 + 107/210 +2021/3960...

7/12>1/2 , 107/210>1/2 and 2021/3960 > 1/2. Thus, this series diverges slowly.

Venture HI - 4 years, 10 months ago

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Hi Nucky,

The idea an infinite product of terms all greater than 1 implies that the series diverges turns out not to be true. A famous example is k=1(1+1ak)=e\prod_{k=1}^{\infty}(1 + \frac{1}{a_k}) = e where a1=1a_1 = 1 and ak+1=(k+1)(ak+1) a_{k+1} = (k+1)(a_k +1) for all k>1.k > 1. I tried proving the harmonic series diverges in a similar way once and had the same assumption!

Roberto Nicolaides - 4 years, 6 months ago

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