$\ln(n+a)-\ln(n-a)=2(\frac an+\frac{a^3}{3n^3}+\frac{a^5}{5n^5}+\dots)$

Question

Show that $$\ln(n+a)-\ln(n-a)=2\left(\frac an+\frac{a^3}{3n^3}+\frac{a^5}{5n^5}+\dots\right)$$ I tried writing $$\ln(n+a)-\ln(n-a)=\ln\frac{n+a}{n-a}=\ln\left(1+\frac{2a}{n-a}\right)$$ and then applying the Maclaurin series of $\ln(1+x)$ but couldn't come up with anything.

Answer 1

Start with $$ \ln (n+a) = \ln n + \ln \left( 1 + \frac an \right) = \ln n + \frac an - \frac 12 \left(\frac an \right)^2 + \frac 13 \left(\frac an \right)^3 - \frac 14 \left(\frac an \right)^4 + \ldots \, , $$ using the MacLaurin series for $\ln (1+x)$. Then do the same for $\ln (n-a)$ and combine the results.

Note that you need $n > 0$ and $|a| < n$ for these developments.

Answer 2

We know that $$\log(n+a)-\log(n-a)=\log n+\log\left(1+\frac an\right)-\log n-\log\left(1-\frac an\right)=\\=\log\left(1+\frac an\right)-\log\left(1-\frac an\right)$$ Then using Taylor series we have: $$\sum_{k=1}^\infty\frac{(-1)^{k+1}a^k}{kn^k}-\sum_{k=1}^\infty\frac{(-1)^{k+1}(-a)^k}{kn^k}=\sum_{k=1}^\infty\frac{(-1)^{k+1}a^k}{kn^k}+\sum_{k=1}^\infty\frac{a^k}{kn^k}=\color{red}{\sum_{k=0}^\infty\frac{a^{k+1}}{(2k+1)n^{2k+1}}}$$