9. Which of the following series converge?
I. $\sum_{n=1}^{\infty} \frac{8^{n}}{n!}$
II. $\sum_{n=1}^{\infty} \frac{n!}{n^{100}}$
III. $\sum_{n=1}^{\infty} \frac{n+1}{(n)(n+2)(n+3)}$
(A) I only
(B) II only
(C) III only
(D) I and III only
(E) I, II, and III

Question

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ For series I, we have $ \sum_{n=1}^{\infty} \frac{8^{n}}{n!} $. This is a standard exponential series which converges to $ e^8 - 1 $. ‖ ‖ step 2 ⋮ For series II, we have $ \sum_{n=1}^{\infty} \frac{n!}{n^{100}} $. This series diverges because the factorial in the numerator grows much faster than the polynomial in the denominator. ‖ ‖ step 3 ⋮ For series III, we have $ \sum_{n=1}^{\infty} \frac{n+1}{(n)(n+2)(n+3)} $. This series converges as shown by the Asksia-LL calculator result. ‖ ∻Answer∻ ***D*** ∻Key Concept∻ ⚹Convergence of Series⚹ ∻Explanation∻ ⚹Series I converges because it is an exponential series, and Series III converges based on the given calculation. Series II diverges due to the rapid growth of the factorial in the numerator.⚹◊What is the convergence test that can be applied to series to determine if they converge or diverge?⍭ Generate me a similar question◊

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