Evaluate $ \lim_{n\to \infty} ( \lim_{x\to0} (1+\tan^2(x)+\tan^2(2x)+ \cdots + \tan^2(nx)))^{\frac{1}{n^3x^2}} $

Question

$ \lim_{n\to \infty} ( \lim_{x\to0} (1+\tan^2(x)+\tan^2(2x)+ \cdots + \tan^2(nx)))^{\frac{1}{n^3x^2}} $

The answer should be $ {e}^\frac{1}{3} $

I haven't encountered problems like this before and I'm pretty confused, thank you.

I guess we must use the remarkable limit of $ \frac{\tan(x)}{x} $ when $ x $ approaches $ 0 $ by dividing and then multiplaying with $x$.

Answer 1

Note that for $x\to 0$

• $\tan x= x+o(x^2)$

then

$$(1+\tan^2(x)+\tan^2(2x)+ \cdots + \tan^2(nx)))^{\frac{1}{n^3x^2}} = (1+x^2+4x^2+...+n^2x^2+o(x^2)) ^ {\frac{1}{n^3x^2}}=\large{e^{\frac{\log(1+x^2+4x^2+...+n^2x^2+o(x^2))}{n^3x^2}}=e^{\frac{x^2+4x^2+...+n^2x^2+o(x^2)}{n^3x^2}}=e^{\frac{n(2n+1)(n+1)}{6n^3}+o(1)} \stackrel{x\to0}\to e^{\frac{n(2n+1)(n+1)}{6n^3}}}$$

and for $n\to \infty$

$$\Large{e^{\frac{n(2n+1)(n+1)}{6n^3}}}\to e^\frac13$$

Answer 2

Here is a sketch. First write

$$\left(1+\sum_{k=1}^n \tan^2(kx) \right)^{1/n^3x^2}=e^{\frac{1}{n^3x^2}\log(1+\sum_{k=1}^n\tan^2(kx))} $$

Now, find the limit as $x\to 0$ exploiting the continuity of the exponential function. For example, L'Hospital's Rule can be applied twice. You may also use asymptotic equivalents (or Taylor's Theorem) if you are familiar with those.

You will arrive at the sum $\frac1{n^3}\sum_{k=1}^nk^2$ as the exponent.

Can you finish now?