Question
What is wrong with the convergence check for $\sum \frac{x^n}{(n+1)!}$?
Original question: How is this wrong
⑥
=
Radius of convergence R=1
-1<x<1 x=-1
it converges because of alternating series test
x=1
converges alt series test
Interval of Convergence
Uh how is the limit zero here?
Expert Verified Solution
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Key takeaway: This is a classic place where a tiny indexing slip changes the whole conclusion. The ratio test is fine; the trouble is what happens after that, especially at the endpoints.
The main issue is that the work mixes up the ratio test result and the endpoint behavior.
We start with
Step 1: Ratio test
Let
Then
=\left|\frac{x^{n+1}}{(n+2)!}\cdot\frac{(n+1)!}{x^n}\right| =\frac{|x|}{n+2}.$$ Now take the limit: $$\lim_{n\to\infty} \frac{|x|}{n+2}=0.$$ That is **0 for every real $x$**, so the series converges for all $x$. ### Step 2: Radius and interval Since the ratio test gives convergence for every $x$, the radius of convergence is $$\boxed{R=\infty}.$$ So the interval of convergence is $$\boxed{(-\infty,\infty)}.$$ ### What went wrong in the shown work? - The conclusion **$R=1$** is incorrect. The limit is not $|x|$; it is $0$. - The endpoint checks at $x=\pm 1$ are unnecessary once you know the radius is infinite. - At $x=1$, the series is $$\sum_{n=0}^{\infty} \frac{1}{(n+1)!},$$ not $\sum 1/n$. So the harmonic-series step is just an indexing mistake. --- **Pitfalls the pros know** 👇 Two traps show up a lot here. First, people cancel factorials incorrectly and lose the extra $n+2$ in the denominator. Second, they rewrite the $x=1$ case as a harmonic series by accident, which is not the same series at all. Once the factorial is tracked correctly, the series converges for every $x$. **What if the problem changes?** If the denominator were only $(n+1)$ instead of $(n+1)!$, then the same ratio test would no longer give convergence for all $x$. The factorial is the reason the terms shrink so quickly. `Tags`: ratio test, index shift, radius of convergence