Which convergence test to use I need to prove that the following converges: 
$$\sum\limits_{n=0}^{\infty} n^3e^{-n^4}$$
I'm not sure which test to use. I've tried ratio test but it gets messy. I don't think I have any sums to use in the comparison test and the integral test obviously wouldn't work. Thanks
 A: Ratio test works fine. The ratio in question is $$\frac{(n+1)^3 e^{-(n+1)^4}}{n^3 e^{-n^4}} = \left( \frac{n+1}{n} \right)^3 e^{n^4-(n+1)^4} = \left( \frac{n+1}{n} \right)^3 e^{-1-4n -6n^2 -4n^3} $$
The exponential term tends to $0$ as $n \to \infty$; the fraction tends to $1$; so the ratio of the terms tends to $0$.

Integral test works fine: $$\int_1^{\infty} x^3 e^{-x^4} dx = \int_1^{\infty}\frac{1}{4} \dfrac{d}{dx}(e^{-x^4}) dx = \frac{1}{4e}$$
To verify that the integrand is decreasing for sufficiently large $x$, differentiate it, obtaining $$3x^2 e^{-x^4} - 4 x^6 e^{-x^4} = x^2 e^{-x^4}(3-4x^4)$$
clearly negative for $x > 1$.

Comparison test works fine: since $e^{-n^4} \leq \frac{1}{n^5}$, have $$n^3 e^{-n^4} \leq n^{-2}$$
It is true that $e^{-n^4} \leq \frac{1}{n^5}$, because that's equivalent to $$n^5 e^{-n^4} \leq 1$$
Differentiating the left-hand side, obtain $$n^4(e^{-n^4}) (5-4n^4)$$
which is negative for integer $n > 1$; so the maximum of the function $n^5 e^{-n^4}$ over the integers is attained at $n=1$, and then we do indeed get $e^{-1} \leq 1$.
A: Directly $\;n-$th root test:
$$\sqrt[n]{n^3e^{-n^4}}=\frac{\left(\sqrt[n]n\right)^3}{e^{n^3}}\xrightarrow[n\to\infty]{}0<1$$
and thus the series converges
A: The form of the term seems to ask for the Integral Test since
$$
\int_x^\infty t^3e^{-t^4}\,\mathrm{d}t=\tfrac14e^{-x^4}
$$
A: First note that
$$\sum\limits_{n=0}^{\infty} \left|\frac{n^3}{e^{n^4}}\right|=\sum\limits_{n=1}^{\infty} \left|\frac{n^3}{e^{n^4}}\right|\leq\sum\limits_{n=1}^{\infty} \left|\frac{n^3}{n^5}\right|=\sum\limits_{n=1}^{\infty} \left|\frac{1}{n^2}\right|$$
Therefore
$$\sum\limits_{n=0}^{\infty} \frac{n^3}{e^{n^4}}=\mbox{absolutely convergent}$$
