Find the interval of convergence of the series [math: ∑n=1∞n!2n(x−5)n] \displaystyle\sum\limits_{n=1}^ \infty \frac{n!}{2^n} (x-5)^n .

Question

BlackTom AI · Accepted Answer

Let me restate the problem clearly: we have the power series sum from n=1 to infinity of (n! / 2^n) * (x - 5)^n. To determine the interval of convergence, I'll examine how the terms behave as n grows large using a standard convergence test for power series, the ratio test, which compares consecutive terms.

First, consider the general term a_n = (n! / 2^n) * (x - 5)^n. The ratio test looks at |a_{n+1} / a_n|. Compute:
|a_{n+1} / a_n| = [ (n+1)! / 2^{n+1} * |x-5|^{n+1} ] / [ n! / 2^n * |x-5|^n ] = [(n+1)/2] * |x-5|.

Now analyze the limit as n → ∞ of this ratio. If |x-5| > 0, then (n+1)/2 * |x-5| grows without bound and thus the ratio diverges to infinity, which means the series cannot converge for any x with x ≠ 5. If |x-5| = 0, i.e., x = 5, then every term becomes zero for n ≥ 1, since (x-5)^n = 0. In that case the series trivially converges (it sums to 0).

This suggests the radius of convergence is zero around x = 5, and the only point where convergence occurs is at x = 5 itself. In interval terms, the set of x-values for which the series converges is precisely {5}.

To connect with the given answer form: the interval of convergence collapses to the single point x = 5. If one thinks in terms of endpoints in a larger interval around 5, there is no interval beyond that single point where the series converges. The behavior at x = 5 yields a perfectly convergent (indeed trivial) sum, whereas any other x causes the ratio to exceed 1 for large n, leading to divergence.

Find the interval of convergence of the series [math: ∑n=1∞n!2n(x−5)n] \displaystyle\sum\limits_{n=1}^ \infty \frac{n!}{2^n} (x-5)^n .单项选择题

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