Finding limit for infinite quantities. 
Find $$\lim_{x\rightarrow 0}{\color{red}{x}} \cdot\bigg(\dfrac{1}{1+x^4}+\dfrac{1}{1+(2x)^4}+\dfrac{1}{1+(3x)^4}\cdots\bigg)$$

As terms are written infinitely my intuitions doesn't let to give answer as $0$. So tried calculated that weird sum something like,
$T_n=\dfrac{1}{1+(nx)^4}=\dfrac{1}{(n^2x^2+\sqrt{2}nx+1)\cdot(n^2x^2-\sqrt{2}nx+1)}$ now this not results in telescopic sum, what should I do, is answer $0$, if so then why should we assume that $\color{red}x$ will stay in numerator no matter whatever happens.
Please help.
 A: Yes, you are right, the limit is not zero. This is my hint: for $x>0$,
$$\int_{x}^{\infty}\frac{dt}{1+t^4}=\sum_{k=1}^{\infty}\int_{kx}^{(k+1)x}\frac{dt}{1+t^4}\leq x\sum_{k=1}^{\infty}\frac{1}{1+(kx)^4}\leq \sum_{k=1}^{\infty}\int_{(k-1)x}^{kx}\frac{dt}{1+t^4}=\int_{0}^{\infty}\frac{dt}{1+t^4}.$$
Then take a look at How can I compute the integral $\int_{0}^{\infty} \frac{dt}{1+t^4}$?
A: Making $n = \frac{1}{x}$
$$
\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^n\frac{1}{1+\left(\frac{k}{n}\right)^4}\equiv\int_0^1\frac{d\xi}{1+\xi^4} = \frac{\pi +2 \coth ^{-1}\left(\sqrt{2}\right)}{4 \sqrt{2}}
$$
NOTE
For integration purposes
$$
\frac{1}{1+\xi^4} = \frac{a_1\xi+b_1}{\xi^2+\sqrt 2 \xi + 1}+\frac{a_2 \xi+ b_2}{\xi^2-\sqrt 2 \xi + 1}
$$
so that
$$
\int\frac{d\xi}{1+\xi^4} = \frac{1}{2} \left(-2 \left(a_1+\sqrt{2} b_1\right) \tan ^{-1}\left(1-\sqrt{2} \xi \right)+a_1 \log \left(\xi ^2-\sqrt{2} \xi
   +1\right)-2 \left(a_2-\sqrt{2} b_2\right) \tan ^{-1}\left(\sqrt{2} \xi +1\right)+a_2 \log \left(\xi ^2+\sqrt{2} \xi
   +1\right)\right)+C
$$
