Question

For any positive real number $a$ and for any $n\in N$, the greatest value of
$\frac{a^n}{1+a+a^2....a^{2n}}$ is

• A. $\frac{1}{2n}$
• B. $\frac{1}{2n+1}$
• C. $\frac{1}{2n-1}$
• D. None of the above.

Answer 1

Answer: (B)

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

We know that $A.M.\ge G.M.$

Therefore, $\frac{1+a+a^2+...+a^{2n}}{2n+1}\ge \sqrt[(2n+1)]{(2n+1)1\ast a\ast a^2\ast ...\ast a^{2n}}$

$⟹\frac{1+a+a^2+....+a^{2n}}{2n+1}\ge \sqrt[(2n+1)]{(2n+1)a^{(1+2+...+2n)}}$

We know that sum of first $n$ numbers is $1+2+...+n=\frac{n(n+1)}{2}$

Therefore $1+2+...+2n=\frac{2n(2n+1)}{2}=n(2n+1)$

$⟹\frac{1+a+...+a^{2n}}{2n+1}\ge (a^{n(2n+1)}{)}^{\frac{1}{2n+1}}$

$⟹\frac{1+a+...+a^{2n}}{2n+1}\ge a^n$

$⟹\frac{a^n}{1+a+...+a^{2n}}\le \frac{1}{2n+1}$

Therefore the greatest value of $\frac{a^n}{1+a+...+a^{2n}}$ is $\frac{1}{2n+1}$