Let S: (-1,1) approaching R... compute S? Maths help

Let S: (-1,1) → R (real numbers) be defined by S(x)= ∑(n=1 to inf.) (-1)^(n-1) (x^n)/n. Compute S'(x) and deduce that, for any x∈ (-1,1), ln(1+x)=∑(n=1 to inf.) (-1)^(n-1) (x^n)/n

Please explain

Thanks

You are probably supposed to recall, or assume, that a power series can be differentiated "term by term" (ie, to compute S'(x), it is enough to compute, for each n, the derivative of (-1)^(n-1) (x^n)/n, and then add all the results up. If you do that you find, using the power rule, that

S'(x) = sum_{n = 1 to infinity} (-1)^(n-1) [n x^(n-1)]/n
= sum_{n = 1 to infinity} (-1)^(n-1) x^(n-1)
= sum_{n = 1 to infinity} (-x)^(n-1).

The point of this calculation is that it shows that S'(x) is represented by a geometric series with first term a = (-x)^(1 - 1) = (-x)^0 = 1, and common ratio of terms r = -x, so its sum is a/(1 - r) = 1/(1 - (-x)) = 1/(1 + x). So

S'(x) = 1/(1 + x) by the geometric series formula.

You are now supposed to recall that if f(x) denotes the function f(x) = ln(1 + x), then f'(x) = 1/(1 + x) also. It follows that [S(x) - f(x)]' = S'(x) - f'(x) = 0 for all x in (-1,1), so that the difference S(x) - f(x) is constant. So there is a number c with S(x) - ln(1 + x) = c, and hence there is some c with

S(x) = c + ln(1 + x) for all x in (-1,1).

If you evaluate both sides at x = 0, you find (just by plugging x = 0 into the series definition of S(x)) that the left hand side is 0, and the right hand side is c + ln(1 + 0) = c + ln(1) = c. It follows that c = 0 and hence

S(x) = ln(1 + x) for all x in (-1,1).