# Prove $\int_\limits{0}^{\infty} n^4 t^2e^{-2nt} dt$ divergence

Show that $x_n(t)=n^2 te^{-nt}$ does not converge in $L_2(\mathbb{R_+})$.

$\int_\limits{0}^{\infty}(n^2 te^{-nt})^2dt=\int_\limits{0}^{\infty} n^4 t^2e^{-2nt} dt\geqslant\int_\limits{0}^{\infty} n^2 t^2e^{-2nt} dt$.

I tried to find a smaller integral that would diverge but I cannot lower the value more I cannot increase.

Question:

How should I prove the convergence?

• That integral clearly converges for all $n$. I think you are supposed to show that the sequence of functions $(x_n)$ does not have a limit in $L^2(\Bbb{R}_+)$. In other words, that there is no function $f\in L^2$ such that $||x_n-f||\to0$ as $n\to\infty$. – Jyrki Lahtonen May 22 '18 at 17:31
Doing the substitution $nt=x$ and $n\,\mathrm dt=\mathrm dx$, you get the integral$$n\int_0^{+\infty}x^2e^{-2x}\,\mathrm dx.$$But the integral $\int_0^{+\infty}x^2e^{-2x}\,\mathrm dx$ converges to some number greater than $0$ and therefore your sequence of integrals diverges.
which goes to $+\infty$ as $n\to\infty$.