Test the uniform convergence of sum from n=1 to infinity
Test the uniform convergence of $\displaystyle\sum_{n=1}^{\infty}\frac{\sqrt{1+2^nx}}{n!}$ on $x\in[0,\infty)$.
Explanation of Method
On any bounded interval $[0,a]$, factorial decay easily dominates the exponential growth of $\sqrt{1+2^nx}$, giving a convergent Weierstrass majorant. On the full unbounded domain $[0,\infty)$, however, each individual term $u_n(x)$ is itself unbounded as $x\to\infty$ — and since uniform convergence of a series forces its terms to tend to $0$ uniformly, an unbounded term at every $n$ rules out uniform convergence on $[0,\infty)$ immediately.
- Bound each term on $[0,a]$ and apply the Weierstrass M-test there.
- Observe that each term is unbounded as $x\to\infty$ on $[0,\infty)$.
- Conclude uniform convergence fails on $[0,\infty)$ via the necessary condition $\sup_x|u_n(x)|\to0$.
Step 1 · Uniform convergence on $[0,a]$
For $x\in[0,a]$: $\sqrt{1+2^nx}\le\sqrt{1+2^na}\le\sqrt{2^na}+1=2^{n/2}\sqrt a+1$ (using $\sqrt{s+t}\le\sqrt s+\sqrt t$). So $$ \frac{\sqrt{1+2^nx}}{n!} \le \frac{2^{n/2}\sqrt a+1}{n!} = \sqrt a\cdot\frac{(\sqrt2)^n}{n!}+\frac1{n!} =: M_n. $$ Since $\sum(\sqrt2)^n/n!=e^{\sqrt2}-1$ and $\sum1/n!=e-1$ both converge, $\sum M_n<\infty$. By the Weierstrass M-test, the series converges uniformly on $[0,a]$ for every fixed $a>0$.
Step 2 · Each term is unbounded on $[0,\infty)$
For any fixed $n\ge1$, as $x\to\infty$: $$ u_n(x)=\frac{\sqrt{1+2^nx}}{n!} \to \infty. $$ So $\sup_{x\ge0}u_n(x)=\infty$ for every $n$.
Step 3 · Non-uniform convergence on $[0,\infty)$
A necessary condition for $\sum u_n(x)$ to converge uniformly on a set $E$ is that $u_n\to0$ uniformly on $E$, i.e. $\sup_E|u_n|\to0$. Here $\sup_{x\ge0}u_n(x)=\infty$ for every $n$, so this necessary condition fails outright — the series does not converge uniformly on $[0,\infty)$.
Summary Table
| Domain | Uniform convergence? |
|---|---|
| $[0,a]$, any fixed $a>0$ | Yes (Weierstrass M-test) |
| $[0,\infty)$ | No ($\sup_x u_n(x)=\infty$ for every $n$) |
Additional Notes
- This gives the cleanest possible proof of non-uniform convergence on an unbounded domain: rather than estimating a remainder, it suffices to note that even a single term fails to be uniformly small — a sharper and more direct obstruction than the boundary-concentration failures seen in Problems 195 and 199.
- The bounded-interval result mirrors Problem 183's pattern ($(1+x/n)^n\to e^x$, uniform on bounded intervals but not on all of $\mathbb R$): factorial or exponential-rate majorants routinely rescue convergence on compact sets while unbounded domains remain out of reach.
- Despite the failure of uniform convergence on $[0,\infty)$, the series still converges pointwise for every fixed $x\ge0$, since $n!$ eventually dominates any fixed exponential rate $2^{n/2}\sqrt x$ — the ratio test confirms this individually at each $x$.
Result
Uniformly convergent on every bounded $[0,a]$, but not on $[0,\infty)$ itself: