so firstly

now we know that \dfrac 1 {1+x}=\dfrac 1 {1 - (-x)}=\displaystyle\sum_{n=0}^{\infty}(-1)^nx^n as |x|<1 (remember x will only range between the limits of the integral), by the formula for the infinite sum of a geometric progression.

substituting this into the integral, we have:

\therefore we can conclude \ln2 = 1-\dfrac 1 2 + \dfrac 1 3 - \dfrac 1 4 + \ldots

(*) you can interchange the integral and sum as long as both the sum and resulting integral converge.

fwiw convergence is not quite sufficient for the interchange. Uniform convergence [and integrability of each individual term in the sum but this is â€śobviousâ€ť since theyâ€™re all continuous] however is, so if you wanted to justify it rigorously (obviously beyond the scope of this question), Iâ€™d note that since Taylor series converge uniformly on compact (in \mathbb R^n this just means closed and bounded) subsets of their radius of convergence, we have \displaystyle \sum_{n = 0}^\infty (-1)^n x^n converges uniformly on [0,t] for all 0 \le t < 1, so:

\begin{align*}\int_0^t \left(\sum_{n = 0}^\infty (-1)^n x^n\right) \mathrm dx & = \sum_{n = 0}^\infty \left(\int_0^t (-1)^n x^n \mathrm dx\right) \\ & = \sum_{n = 0}^\infty \frac {(-1)^n} {n + 1} t^{n + 1} \\ & = \ln(1 + t)\end{align*}

taking t \to 1^- gives the result. (https://en.wikipedia.org/wiki/Abel's_theorem)

Donâ€™t worry if this is all analytic gobbledegook at this point, just thought Iâ€™d chip in! This is the type of thing youâ€™d see in the second year.

This is really valuable!!

v cool lol xD

i like analysis, proves so many intuitive results rigorously