Using the Weierstrass test, show that the series sum

Using the Weierstrass test, show that the series $\displaystyle\sum_{n=1}^{\infty}\frac1{x^2+n^3}$ converges uniformly on $(-\infty,\infty)$.

Step 1 · Uniform bound

Since $x^2\ge0$: $x^2+n^3\ge n^3$, so $\dfrac1{x^2+n^3}\le\dfrac1{n^3}$ for every $x\in\mathbb R$.

Step 2 · Majorant series

$\sum_{n=1}^{\infty}1/n^3=\zeta(3)\approx1.202<\infty$ ($p$-series, $p=3>1$).

Step 3 · M-test and continuity

With $M_n=1/n^3$, both M-test conditions hold, so the series converges uniformly and absolutely on $\mathbb R$. Since each term is continuous in $x$, the uniform limit $f(x)=\sum_{n=1}^{\infty}1/(x^2+n^3)$ is continuous on all of $\mathbb R$.

Summary Table

Additional Notes

• This is the exact uniform-convergence upgrade of Problem 169's pointwise result (which used the exponent $2$ instead of $3$): both series are dominated by a convergent $p$-series independent of $x$, so both automatically satisfy the Weierstrass M-test with no restriction on $x$ at all.
• All terms are non-negative, so — as in Problem 169 — absolute convergence and ordinary convergence coincide trivially; the only new content here relative to Problem 169 is the uniformity of the convergence, which the M-test delivers for free.
• As $|x|\to\infty$, the sum $f(x)\to0$ (each term individually vanishes, and the tail is uniformly small by the same $M_n$ bound), giving $f$ the qualitative shape of a bump function peaked at $x=0$.