Find derivative of $f(x)=\tan^2x \sqrt{\tan x\sqrt[3]{\tan x\sqrt[4]{\tan x...}}}$

Question

Find $f(x)$ for $f(x)=\tan^2x \sqrt{\tan x\sqrt[3]{\tan x\sqrt[4]{\tan x...}}}$

I found that the right expression is an infinite series of $\sum^{\infty}_{n=2}{\frac{1}{n!}}$. I know the series converge but how can I evalute $\sum^{\infty}_{n=2}{\frac{1}{n!}}$?

Any hint or idea will be appreciated. Thank you!

Answer 1

$$f(x)=\tan^2x \sqrt{\tan x\sqrt[3]{\tan x\sqrt[4]{\tan x...}}}=\tan^2 x\cdot \left(\tan x\right)^{\sum _{n=2}^{\infty } \frac{1}{n!}}$$

We have $$\sum _{n=2}^{\infty } \frac{1}{n!}=e-1-1=e-2$$

$$f(x)=\tan^2 x\cdot \left(\tan x\right)^{e-2}=\left(\tan x\right)^e$$

Derivative is $$f'(x)=e \,\frac{\tan ^{e-1} x}{ \cos ^2 x}$$