But then if you integrate the infinite sum term by term, you'll get:
\int (1 - x + x^2 - x^3 + ...) = x - x^2/2 + x^3/3 - ...
On the other hand, if you integrate 1/(1+x), you'll get precisely log(1+x). Now, one would want to argue that:
log(1+x) = x - x^2/2 + x^3/3 - ...
and here this is actually true, but in general this may not be true that the integral of the sum of infinite series is equal to sum of the integrals of each term -- what you need here is the notion of uniform convergence, but fortunately 1 - x + x^2 - ... converges uniformly.
But, since Leonard Euler wouldn't really be pedantic about stuff like this, as long as the result is correct, neither should the occasional math hacker. Thus, we put x = 1 in both sides of the equality and we get 1 - 1/2 + 1/3 - ... = log(2).
Apart from integrating term by term, there's one more problem with above reasoning, and there are bonus points for people who notice that (hint: uggc://ra.jvxvcrqvn.bet/jvxv/Nory'f_gurberz )
High school algebra tells us that
1 - x + x^2 - ... + (-1)^n x^n = 1 + (-x) + (-x)^2 + ... + (-x)^n = (1 - (-x)^{n+1})/(1-(-x)).
Thus, for the infinite sum we have:
1 - x + x^2 - x^3 + ... = lim_{n -> inf} (1 - x + x^2 - ... + (-1)^n x^n) = lim_{n -> inf} (1 - (-x)^{n+1})/(1-(-x)) = 1/(1+x).
But then if you integrate the infinite sum term by term, you'll get:
\int (1 - x + x^2 - x^3 + ...) = x - x^2/2 + x^3/3 - ...
On the other hand, if you integrate 1/(1+x), you'll get precisely log(1+x). Now, one would want to argue that:
log(1+x) = x - x^2/2 + x^3/3 - ...
and here this is actually true, but in general this may not be true that the integral of the sum of infinite series is equal to sum of the integrals of each term -- what you need here is the notion of uniform convergence, but fortunately 1 - x + x^2 - ... converges uniformly.
But, since Leonard Euler wouldn't really be pedantic about stuff like this, as long as the result is correct, neither should the occasional math hacker. Thus, we put x = 1 in both sides of the equality and we get 1 - 1/2 + 1/3 - ... = log(2).
Apart from integrating term by term, there's one more problem with above reasoning, and there are bonus points for people who notice that (hint: uggc://ra.jvxvcrqvn.bet/jvxv/Nory'f_gurberz )