Question

Math

Posted 4 months ago

```
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
```

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Answer from Sia

Posted 4 months ago

Solution by Steps

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.

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