Suppose the power series $\displaystyle \sum_{n=1}^{\infty}a_nx^n$ has radius of convergence 4, then the series $\displaystyle \sum_{n=1}^{\infty}a_n2^n$

Question

BlackTom AI · Accepted Answer

Radius of convergence 4 means the power series ∑ a_n x^n converges absolutely for every x with |x| < 4 and diverges for every x with |x| > 4; the behavior at |x| = 4 may be indeterminate. A standard characterization is R = 1 / (limsup_{n→∞} |a_n|^{1/n}).

Step-by-step reasoning:
1. Let L = limsup_{n→∞} |a_n|^{1/n}. Since the radius R = 4, we have
   R = 1/L, so L = 1/4.
2. Consider the series ∑ a_n 2^n. Apply the root test (or use the limsup formula):
   limsup_{n→∞} |a_n 2^n|^{1/n} = limsup_{n→∞} |a_n|^{1/n} · 2 = L · 2.
3. Substitute L = 1/4 to get
   limsup_{n→∞} |a_n 2^n|^{1/n} = (1/4) · 2 = 1/2.
4. Because this limsup is 1/2 < 1, the root test implies the series ∑ a_n 2^n converges absolutely, and hence converges.

Why the correct answer is correct:
- Absolute convergence for x inside the radius is guaranteed by the definition of radius of convergence. Here |2| = 2 < 4, so absolute convergence follows. The quantitative root-test calculation above confirms this.

Why the other choices are incorrect:
- b. diverges is false because the point x = 2 lies inside the open interval of convergence (|x| < 4), so divergence cannot occur for every choice of coefficients consistent with R = 4.
- c. may converge or diverge depending on a_n is false because the radius information alone guarantees convergence at every x with |x| < 4, independent of the particular a_n (provided they produce R = 4).

Conclusion: The series ∑ a_n 2^n converges, matching answer a. converges.

Suppose the power series \(\displaystyle \sum_{n=1}^{\infty}a_nx^n\) has radius of convergence 4, then the series \(\displaystyle \sum_{n=1}^{\infty}a_n2^n\)单项选择题

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