Question

Applying the Ratio Test with \( a_{n}=\frac{|x|^{n}}{n !} \) shows that the series \( \sum_{n=1}^{\infty} \frac{x^{n}}{n !} \) a) converges absolutely for all \( x \in \mathbb{R} \). b) converges only if \( x=0 \). c) converges absolutely for all \( x \) with \( |x|<1 \) and diverges if \( |x|=1 \). d) None of the above.

          Applying the Ratio Test with \( a_{n}=\frac{|x|^{n}}{n !} \) shows that the series \( \sum_{n=1}^{\infty} \frac{x^{n}}{n !} \)
a) converges absolutely for all \( x \in \mathbb{R} \).
b) converges only if \( x=0 \).
c) converges absolutely for all \( x \) with \( |x|<1 \) and diverges if \( |x|=1 \).
d) None of the above.

Applying the Ratio Test with an=(|x|^n)/(n !) shows that the series ∑n=1^∞(x^n)/(n !)
a) converges absolutely for all x ∈ℝ.
b) converges only if x=0.
c) converges absolutely for all x with |x|<1 and diverges if |x|=1.
d) None of the above.

Added by Jing W.

Calculus: Early Transcendentals

James Stewart 8th Edition

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Solved by Expert Narayan Hari

Step 1

The Ratio Test states that for a series \(\sum_{n=1}^{\infty} a_n\), if the limit \[ L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \] exists, then the series converges absolutely if \(L < 1\), the test is inconclusive if \(L = 1\), and the series Show more…

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