Using the Weierstrass test, show that the series sum

Using the Weierstrass test, show that the series $\displaystyle\sum_{n=1}^{\infty}\frac1{2^{n-1}\sqrt{1+nx}}$ converges uniformly on $[0,\infty)$.

Step 1 · Uniform bound

For $x\ge0$ and $n\ge1$: $1+nx\ge1$, so $\sqrt{1+nx}\ge1$, giving $$ \frac1{2^{n-1}\sqrt{1+nx}} \le \frac1{2^{n-1}} $$ for every $x\in[0,\infty)$.

Step 2 · Majorant series

$\displaystyle\sum_{n=1}^{\infty}\frac1{2^{n-1}} = 1+\frac12+\frac14+\cdots = 2$, a convergent geometric series with ratio $1/2$.

Step 3 · Weierstrass M-test

With $M_n=1/2^{n-1}$, both M-test conditions hold, so the series converges uniformly and absolutely on $[0,\infty)$. Since each term is continuous in $x$ on this domain, the sum is continuous on $[0,\infty)$ as well.

Summary Table

Additional Notes

• This is the fastest-converging majorant series among the M-test problems in this batch aside from Problem 193's factorial bound — the geometric decay $1/2^{n-1}$ needs no help at all from the $\sqrt{1+nx}$ factor, which only makes convergence faster still as $x$ grows.
• As $x\to\infty$, each individual term $\to0$, and in fact the whole sum $\to0$ by the same uniform bound — the function defined by this series decays to $0$ as $x\to\infty$, unlike Problem 194's bump-shaped limit.
• Restricting to $x\ge0$ is essential only to keep $1+nx\ge1$ cleanly; the same bound (with a different constant) would work on any half-line $x\ge-c$ with $c<1/n$ for all relevant $n$, but $[0,\infty)$ is the natural domain where the square root is always defined and the bound is uniform in $n$.