Let phi: R to R be a bounded function. Using
Let $\varphi:\mathbb R\to\mathbb R$ be a bounded function. Using the Weierstrass test, show that the series $\displaystyle\sum_{n=1}^{\infty}\frac1{n^2+[\varphi(x)]^2}$ converges uniformly on $(-\infty,\infty)$.
Explanation of Method
Since $[\varphi(x)]^2\ge0$ for every $x$, dropping this term from the denominator gives the immediate $x$-independent bound $1/n^2$, and $\sum1/n^2=\pi^2/6$ converges — a direct generalization of Problems 169 and 194, valid for any real-valued $\varphi$ (boundedness of $\varphi$ is not even needed for this particular bound, though it matters for the shape of the resulting sum).
- Bound each term by $1/n^2$, uniformly in $x$.
- Confirm $\sum1/n^2=\pi^2/6$ converges.
- Apply the Weierstrass M-test to conclude uniform convergence.
Step 1 · Uniform bound
For any real number $\varphi(x)$: $[\varphi(x)]^2\ge0$, so $n^2+[\varphi(x)]^2\ge n^2$, giving $$ \frac1{n^2+[\varphi(x)]^2} \le \frac1{n^2} $$ for every $x\in\mathbb R$ — note this bound requires no assumption on $\varphi$ beyond being real-valued.
Step 2 · Majorant series
$\displaystyle\sum_{n=1}^{\infty}\frac1{n^2}=\frac{\pi^2}6<\infty$ ($p$-series, $p=2>1$).
Step 3 · Weierstrass M-test
With $M_n=1/n^2$, both M-test conditions hold, so the series converges uniformly and absolutely on $\mathbb R$, regardless of which (real-valued) function $\varphi$ is used. If $\varphi$ is additionally continuous, the resulting sum is continuous on $\mathbb R$ as a uniform limit of continuous functions.
Summary Table
| Quantity | Value |
|---|---|
| Majorant | $M_n=1/n^2$ |
| $\sum M_n$ | $\pi^2/6$ |
Additional Notes
- This problem is the general template behind Problems 169 and 194: both $x^2$ (Problem 194) and any bounded $[\varphi(x)]^2$ here are just specific non-negative "shift" terms added to $n^2$ or $n^3$ in the denominator, and the same drop-the-shift trick handles all of them uniformly.
- The word "bounded" in the hypothesis is not actually needed for uniform convergence itself (the bound $1/n^2$ works for any real $\varphi$), but it does guarantee the resulting sum $f(x)=\sum1/(n^2+[\varphi(x)]^2)$ is itself a bounded function of $x$, squeezed between $\sum1/(n^2+M^2)$ and $\pi^2/6$ if $|\varphi(x)|\le M$.
- Composing with $\varphi(x)=\sin x$ or $\varphi(x)=\arctan x$ recovers concrete bounded examples in the spirit of Problems 181–183's arctan and exponential sequences, all sharing this same uniform-convergence mechanism.
Result
Converges uniformly and absolutely on $\mathbb R$ for any bounded (indeed any real-valued) $\varphi$: