Harmonic
Harmonic¶
i'm going to start with the alternating harmonic series \(1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5}...\)
then, for example \(- \frac{1}{4} = \frac{1}{4} - \frac{1}{2}\)
then, the original series is equal to \(1 + \frac{1}{2} - 1 + \frac{1}{3} + \frac{1}{4} - \frac{1}{2} ...\)
also, I will call the total sum "?"
now, let me define \(f(n)\) as \(f(1) = 1\) and \(f(2) = 1 + \frac{1}{2} - 1 + \frac{1}{3} = \frac{1}{2} + \frac{1}{3}\) than \(f(3)\) equals the next few terms, so \(\frac{1}{3} + \frac{1}{4} + \frac{1}{5}\)
in genaral, \(f(n)\) = \(\frac{1}{n} + \frac{1}{n + 1} + ...\) with n terms OR \(\frac{1}{n} + \frac{1}{n} + ... + \frac{1}{2n - 1}\)
but than \(f(\infty) = ?\)
\(f(n,x) = \frac{1}{n} + \frac{1}{n + 1} + ... + \frac{1}{xn - 1}\) \(x\) as \(\frac{\text{integer}}{n}\) for integer n
\(f(n,x + \frac{1}{n}) - f(n,x) = \frac{1}{xn}\)
\(n(f(n,x + \frac{1}{n}) - f(n,x)) = \frac{1}{x}\)
as \(n \to \infty\) \(\frac{1}{n} \to 0\) and \(x\) could be anything AND the previous expression equals \(\frac{df(\infty,x)}{dx}\)
but \(ln \prime(x) = \frac{1}{x}\) and \(ln(1) = 0\) AND \(f(\infty,1) = \frac{1}{1 + \infty} = 0\)
therefore \(f(\infty,x) = ln(x)\)
but \(f(n,2) = f(n)\) and \(? = f(\infty)\)
so, \(? = f(\infty,2) = ln(2)\)
so \(1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5}\) FOREVER $ = ln(2)$
yay!
regular harmonic¶
Now I'll start with a seiries like \(1 + \frac{1}{2} + \frac{1}{3} - 1 + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} - \frac{1}{2}\) is \(f(\infty, 3)\), so ln(3)
the simpler \(1 + \frac{1}{2} + \frac{1}{3} - 1 - \frac{1}{2} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} - \frac{1}{3} - \frac{1}{4}\), if you don't get it, its 3 positave 2 negatave.
for the same reasons as before, this equals (remember that \(n \to \infty\) in the following) \(\frac{1}{2n} + \frac{1}{2n + 1} + ... + \frac{1}{3n - 1}\), so \((\frac{1}{n} + \frac{1}{n + 1} + ... + \frac{1}{3n - 1}) - (\frac{1}{n} + \frac{1}{n + 1} + ... + \frac{1}{2n - 1})\), so \(f(n, 3) - f(n, 2)\), so \(f(\infty, 3) - f(\infty, 2)\), so \(ln(3) - ln(2)\), so \(ln(\frac{3}{2})\), now I will define \(g(m, n)\) as m posataves, n negataves, so \(g(m, n) = ln(\frac{m}{n})\)
now, with the power of \(g\) on our side, we can solve for \(1 + \frac{1}{2} + \frac{1}{3}...\) (which by the way is the point of this chapter) as 1 posatave, 0 negataves, so \(ln(1) - ln(0)\), but \(ln(0)\) is undefined, so time to define it!
\(ln(0)\)¶
\(ln(0) = u\) (u for undefined)
\(e^u = 0\)
\(\frac{1}{e^{-u}} = 0\)
\(e^{\infty} = \infty\)
\(\frac{1}{\infty} = 0 \Rightarrow u = - \infty \quad \And \quad ln(1) = 0 \Rightarrow 1 + \frac{1}{2} + \frac{1}{3}... = \infty\)
\(\frac{1}{\infty}\)¶
I'll just say it, I spent a long time trying to figure it out a way to prove that \(1 + \frac{1}{2} + \frac{1}{3}... = \infty\), and I realized that \(e^{-\infty} = 0\) and \(\frac{1}{\infty} = 0\) are true by convention, and the only proofs like "if you divide \(1\) by \(e\) engough times, it eventually has to hit zero" and "\(\frac{1}{\infty}\) 'aproaches ' zero" are nor rigoris because the word "approach" proveably doesent have rigoris definition, so you just have to trust the conclution.