Prove:—
$\log _{e}(n+1)-\log _{e} n=\frac{1}{n}-\frac{1}{2 n^{2}}+\frac{1}{3 n^{3}}-\ldots$
$\log _{e} n-\log _{e}(n-1)=\frac{1}{n}+\frac{1}{2 n^{2}}+\frac{1}{3 n^{3}}+\ldots$
Here's my attempt:—
I know that,
$e^{c y}=1+cy+\frac{c^{2} y^{2}}{2 !}+\frac{c^{3} y^{3}}{3 !}+\ldots$
Putting $e^c=a$ I get,
$a^{y}=1+y \log _{e} a+\frac{y^{2}\left(\log _{e} a\right)^{2}}{2 !}+\frac{y^{3}\left(\log _{e} a\right)^{3}}{3 !}+\ldots$
Putting $1+n=a$,
$(1+n)^{y}=1+y \log _{e}(1+n)+\frac{y^{2}}{2 !}\left\{\log _{e}(1+n)\right\}^{2}+\frac{y^{3}}{3 !}\left\{\log _{e}(1+n)\right\}^{3}+\ldots$
By using Binomial Theorem, when $n<1$.
$(1+n)^{y}=1+y n+\frac{y(y-1)}{2 !} n^{2}+\frac{y(y-1)(y-2)}{3 !} n^{3}+\ldots$
In the above equation the coefficient of $y$ is
$x+\frac{(-1)}{1 \cdot 2} x^{2}+\frac{(-1)(-2)}{1 \cdot 2 \cdot 3} x^{3}+\frac{(-1)(-2)(-3)}{1 \cdot 2 \cdot 3 \cdot 4} x^{4}+\ldots$
By comparing the coefficients of $y$ I get
$\log _{e}(1+n)=n-\frac{x^{2}}{2}+\frac{n^{3}}{3}-\frac{n^{4}}{4}+\ldots$.
Putting $n=a$
$(n)^{y}=1+y \log _{e}(n)+\frac{y^{2}}{2 !}\left\{\log _{e}(n)\right\}^{2}+\frac{y^{3}}{3 !}\left\{\log _{e}(n)\right\}^{3}+\ldots$
Now, here I am able to obtain the expansion (or series) of $\log _{e}(1+n)$ but I am not able to do the same for $\log _{e}(n)$ because I can't obtain the expansion of (n)^y.
Also,
By my attempt it seems like all these equations are applicable for only $n<1$
So,
Is there a more general/elegant proof applicable for all $n∈R$?
Substitute $x=1/n$ into the Taylor series for $ \ln(x)$ \begin{eqnarray*} \ln(1+x) = x-\frac{x^2}{2}+ \frac{x^3}{3}-\cdots. \end{eqnarray*} and use \begin{eqnarray*} \ln(ab) = \ln(a)+ \ln(b). \end{eqnarray*}