Question

$2{\log }_{e}n-{\log }_e(n+1)-{\log }_e(n-1)=\frac{1}{n^2}+\frac{1}{a{n}^4}+\frac{1}{3n^6}+...\infty$
Find $a$.

Answer 1

L.H.S. $=2{\log }_{e}n-{\log }_e(n+1)-{\log }_e(n-1)$

$={\log }_e{n}^2-[\log (n+1)+\log (n-1)]$

$={\log }_e{n}^2-\log (n+1)(n-1)$

$={\log }_e{n}^2-{\log }_e(n^2-1)$

$=-[{\log }_e(n^2-1)-{\log }_e{n}^2]$

$=-{\log }_e\left(\frac{n^2-1}{n^2}\right)$

$={\log }_e(1-\frac{1}{n^2})$

$=-[-\frac{1}{n^2}-\frac{1}{2n^4}-\frac{1}{3n^6}-...]$

$=\frac{1}{n^2}+\frac{1}{2n^4}+\frac{1}{3n^6}+...\infty$

so $a=2.$