A series is defined such that $$a_n = {(2n+1)}^{\frac 1{\sqrt {2n+1}}}, n \in \mathbb Z^+$$
The greatest term of this series is to be found.
I considered defining a function, $f(x) ={(2x+1)}^{\frac 1{\sqrt {2x+1}}}$, taking its derivative and then equating it to zero.
By taking logarithm on both sides of the equation, differentiating and rearranging, I found $$ \frac d{dx} f(x) = \frac {\ln (2x+1)(2-\ln(2x+1))}{(2x+1)^2}$$
Equating this to zero and solving for $x$, I found $$x= \frac{e^2-1}2$$ $$\implies 2x+1 \approx 7.4$$ The options for the greatest term were
A)$3^{\frac 1{\sqrt 3}}$
B)$5^{\frac 1{\sqrt 5}}$
C)$7^{\frac 1{\sqrt 7}}$
D)Series is not bounded.
I believe (C) is the correct answer as it was closest to ${7.4}^{\frac 1 {\sqrt {7.4}}}$ . Is this the correct answer?
Is there any alternate approach?