Taylor Series about the function $\log(x + 1)$

Question

I read the following taylor expresion for $$\log (x+1)=\sum _{k=1}^n \frac{(-1)^{k+1} x^k}{k}+\frac{(-1)^n x^{n+1} \, _2F_1(1,n+1;n+2;-x)}{n+1}$$ Do you know where it comes from?

Answer 1

Hint. One may start with $$ \begin{align} \log(1+x)&=\int_0^1 \frac{x}{1+xt} \, dt \\\\&=x\int_0^1 \frac{1-(-1)^{n}(tx)^{n}}{1+xt} \, dt+\int_0^1 \frac{(-1)^{n}x^{n+1}t^n}{1+xt} \, dt \\\\&=x\int_0^1\sum _{k=1}^n(-xt)^{k-1}\:dt+\int_0^1 \frac{(-1)^{n}x^{n+1}t^n}{1+xt} \, dt \\\\&=\sum _{k=1}^n \frac{(-1)^{k+1} x^k}{k}+(-1)^{n}x^{n+1}\int_0^1 \frac{t^n}{1+xt} \, dt \tag1 \end{align} $$ then one may deduce the desired identity from the Euler result (1748) concerning $_2F_1(a,b;c;z)$: $$ B(b,c-b)\,_2F_1(a,b;c;z) = \int_0^1 t^{b-1} (1-t)^{c-b-1}(1-xt)^{-a} \, dt \qquad \text{Re}(c) > \text{Re}(b) > 0, $$ with $b=n+1$, $c=n+2$, $a=1$, $z=-x$, which gives $$ \frac{\Gamma(n+1)\Gamma(1)}{\Gamma(n+2)}\,_2F_1(1,n+1;n+2;-x) = \int_0^1 \frac{t^{n}}{1+xt} \, dt. \tag2 $$ Inserting $(2)$ into $(1)$ yields the sought identity.

Answer 2

Another was is to note that

$\begin{array}\\ (1+x)\sum_{k=0}^n (-1)^k x^k &=\sum_{k=0}^n (-1)^k x^k+\sum_{k=0}^n (-1)^k x^{k+1}\\ &=\sum_{k=0}^n (-1)^k x^k+\sum_{k=1}^{n+1} (-1)^{k-1} x^{k}\\ &=1+\sum_{k=1}^n ((-1)^k+(-1)^{k-1}) x^k+ (-1)^{n} x^{n+1}\\ &=1+ (-1)^{n} x^{n+1}\\ \end{array} $

so that $\frac1{1+x} =\sum_{k=0}^n (-1)^k x^k-\frac{(-1)^{n} x^{n+1}}{1+x} $.

Integrating from $0$ to $t$,

$\begin{array}\\ \ln(1+t) &=\int_0^t \frac{dx}{1+x}\\ &=\int_0^t \left(\sum_{k=0}^n (-1)^k x^k-\frac{(-1)^{n} x^{n+1}}{1+x}\right)dx\\ &= \sum_{k=0}^n (-1)^k \int_0^tx^kdx-(-1)^n\int_0^t\frac{ x^{n+1}}{1+x}dx\\ &= \sum_{k=0}^n (-1)^k \frac{t^{k+1}}{k+1}-(-1)^n\int_0^t\frac{ x^{n+1}}{1+x}dx\\ \end{array} $

which gives the same integral as before (after scaling to $0$ to $1$).