What makes the natural log "natural"? | Ep. 7 Lockdown live math

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Recommends Elias Zamaria · 2021-02 · Mathematics Stack Exchange · on “Sum of the alternating harmonic series $\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k} = \frac{1}{1} - \frac{1}{2} + \cdots $”

Grant Sanderson, aka 3Blue1Brown, has a good explanation of this [this video] in one of his Lockdown Math videos. His explanation, summarized: $1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \ldots = f(1)$ , where $f(x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \frac{x^5}{5} -\ldots$ $\frac{df}{dx} = 1 - x + x^2- x^3 + x^4 - \ldots = \frac{1}{1+x}$ (when $-1 ≤ x ≤ 1$ ) $f(x) = \ln(1+x)$ $\therefore f(1) = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \ldots = \ln(2)$

Cites Bipasha · 2020-09 · Mathematics Stack Exchange · on “Why does $\sum_{p\in P} \sum_{k\in N^+} \frac {1}{k.^{2}} $ yield $\ln(\frac{\pi^2}{6})$?”

The source for this problem is this 3b1b video [this video] . I understand this: $$\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...=\frac{\pi^2}{6}$$ Now he alters the series to include only primes and powers of primes (eg. 4 and 8 are included because they are powers of 2, which is prime) while scaling down the powers of primes by a factor of the exponent, as in: $$\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2(2)}+\frac{1}{5^2}+\frac{1}{7^2}+\frac{1}{8^2(3)}+\frac{1}{9^2(2)}+...$$ This happens to equate to: $$ \ln\left(\frac{\pi^2}{6}\right)$$ I tried to search the proof of this for a while, but could not find anything. I would be delighted to see an elementary explanation to this, Thanks!

Cites Francisco Jácome Maura · 2021-09 · Mathematics Stack Exchange · on “Alternating sum involving reciprocals of powers of primes equals $\ln(\frac{\pi}{4})$”

I found the following identity in a [this video] : $$-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{2\cdot9}-\frac{1}{11}+\frac{1}{13} -\frac{1}{17}+\frac{1}{19}-\frac{1}{23}+\frac{1}{2\cdot25}-\frac{1}{3\cdot 27}+ \cdots = \ln\left( \frac{\pi}{4} \right).$$ As noted in the video, the identity is apparently related to the Leibniz formula for $\pi$ [LINK] , but I cannot find how it's derived, or even if it's true. I have searched for information, but found nothing about it. I didn't find this question in math.stackexchange neither. I've tried similar method than the used in this problem [LINK] , and also using directly the series of $\ln(x+1)$ , but I have no idea of how to achieve the identity. It

Cites Chris Bedford · 2024-02 · Mathematics Stack Exchange · on “How is $\log _{10}(e)=\left[\log _e(10)\right]^{-1}$?”

I am watching a logarithm lecture from 3Blue1Brown (great for math dummies like me!) Here is the video for context: [this video] The step that I did not follow is when he equates the LHS and RHS of the equation below. Maybe someone can fill me in on the general rule that allows us to claim this equality? thanks in advance ! $\log _{10}(e)=\left[\log _e(10)\right]^{-1}$