Show that $\log_e (n+a)- \log_e (n-a)=2(\frac{a}{n}+\frac{a^3}{3n^3}+\frac{a^5}{5n^5}+\cdots )$

Question

I adopted this problem from higher algebra book by hall and knight pg.no 196 problem no. 3. I can't derive the proof.

Answer 1

$\log_e (1+t)=t-\frac {t^{2}} 2+\frac {t^{3}} 3-...$. Hence

$\log_e (1-t)=-t-\frac {t^{2}} 2-\frac {t^{3}} 3-...$.

This gives

$\log_e (1+t)-\log_e(1-t)=2[t+\frac {t^{3}} 3+\frac {t^{5}} 5+...]$.

Put $t =\frac a n$ this and observe that $$\log_e (1+\frac a n)-\log_e(1-\frac a n)$$ $$=[\log_e(n+a)-\log_e(n)] -[\log_e(n-a)-\log_e(n)]$$ $$=\log_e(n+a)-\log_e(n-a).$$