Question

The sum of the series $\left( 1+\frac{{{({{\log }_{e}}\,n)}^{2}}}{2!}+\frac{{{({{\log }_{e}}n)}^{4}}}{4!}+... \right)$ is

• A. $n+\frac{1}{n}$
• B. $\frac{n'}{2}+\frac{1}{2n}$
• C. ${{\log }_{e}}\frac{1}{1-{{({{\log }_{e}}n)}^{2}}}$
• D. $\frac{1}{2}{{\log }_{e}}\,\frac{1}{1-{{({{\log }_{e}}\,n)}^{2}}}$

Answer 1

Answer: (B)

$\frac{n'}{2}+\frac{1}{2n}$

Answer 2

We know that, $1+\frac{{{x}^{2}}}{2!}+\frac{{{x}^{4}}}{4!2}+....=\frac{{{e}^{x}}+{{e}^{-x}}}{{}}$
$\therefore$ $a+\frac{{{({{\log }_{e}}n)}^{2}}}{2!}+\frac{{{({{\log }_{e}}n)}^{4}}}{4!}+....$
$=\frac{{{e}^{{{\log }_{e}}n}}+{{e}^{-{{\log }_{e}}n}}}{2}$
$=\frac{n+\frac{1}{n}}{2}$
$=\frac{n}{2}+\frac{1}{2n}$