I have a problem in expressing ln (1 + x) as a convergent series. Can people show me the steps of doing this? thanks in advance!!!
You could expand ln(1+x) in its Taylor series around x=0.
For this, you need all the derivative of ln(1+x) evaluated at x=0.
Let f(x)=ln(1+x). (note first that f(0) = 0).
Then
f'(x) = 1/(1+x).
f''(x) = -1/(1+x)^2.
f'''(x) = 2/(1+x)^3.
f''''(x) = -6/(1+x)^4
and so on.
In general, we have that the n'th derivative of f is $\displaystyle (-1)^{n+1}\frac{(n-1)!}{(1+x)^n}$ (which you may prove by induction). Then
$\displaystyle f^{(n)}(0) = (-1)^{n+1}(n-1)!,$
and so the Taylor series is:
$\displaystyle \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}x^n,$
which converges for $\displaystyle |x|\leq 1$ (except for $\displaystyle x=-1$).
Thanks! Your response is extremely lucid and helpful! It does not only answer the question perfectly, but it also gives me some very good intuition to understand Taylor series expansion! I have become more confident in real analysis now!
Thank you very much
$\displaystyle \displaystyle \frac{1}{1+x} = \frac{1}{1-(-x)} = 1 - x + x^2 - x^3 + ...
$
integrate ...
$\displaystyle \displaystyle \ln(1+x) = C + x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + ...$
$\displaystyle x = 0 \implies C = 0$
$\displaystyle \displaystyle \ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + ... = \sum_{n=1}^{\infty} \frac{(-1)^{n+1} x^n}{n}$