S+xS&=1\\ S&=\frac {1} {1+x}\\ \end {align} To prove in the other direction, use the binomial theorem or simply compute the series about $0$ manually.

Apr 19, 2019 · 5 I was just wondering where the minus sign in the first term of the Taylor expansion of $ \ln (1-x) $ comes from?

Jul 28, 2015 · For which $x$ do you want to prove the inequality? $\ln (1+x)$ is not defined for $x\le -1$, the inequality is false for $x=0$.

I saw this in a proof for the Central Limit Theorem: $\\ln(1-x) \\approx -x$ when $x$ is small It seems to be true when I plug in small values of $x$. But why does it ...

Sep 19, 2016 · Series Expansion for $\ln (x)$ Ask Question Asked 9 years, 3 months ago Modified 2 years, 8 months ago

Sep 14, 2024 · That is, $$ \ln (1+x) = -\sum_ {n=1}^\infty (-1)^n \frac {1} {n} x^n, x \in (-1,1] \text {.} $$ There is a third question: just because a series recipe converges does not mean that the …

Sep 21, 2015 · Looking for Taylor series expansion of $\ln (x)$ Ask Question Asked 10 years, 3 months ago Modified 2 years, 7 months ago

Aug 31, 2019 · Try proving $F (x)=\ln {x}-x+1$ is negative, by using first derivative test to conclude about where $F$ is increasing and where it's decreasing.

Nov 6, 2019 · Put another way, the function $\log (1 + x)$ has one derivative, $\frac {1} {1 + x}$, and that latter function has many antiderivatives, but only one of them is the function we …

Oct 19, 2017 · Since differentiation term by term is allowed inside the interval of convergence and the derivative term by term of the given series is $$ \sum_ {n\ge1} (-1)^nx^ {n-1}=\frac {1} …