Given that f ( n ) ( 0 ) = ( n + 1 ) ! for n = 0 , 1 , 2 , … , which of the following is the Maclaurin series for f ( x ) ?

f ( x ) = ∑ n = 0 ∞ ( n + 1 ) ! x n ( n + 1 ) !

f ( x ) = ∑ n = 0 ∞ n ! x n n !

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1. Limits and Continuity2h 2m
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1. Limits and Continuity

Introduction to Limits

Calculus

1. Limits and Continuity

Introduction to Limits

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Multiple Choice

Given that $f^{(n)} (0)$ = $(n + 1)!$ for $n = 0, 1, 2, \dots$ , which of the following is the Maclaurin series for $f (x)$ ?

f (x) = \sum_{n = 0}^{\infty} (n + 1)! \frac{x^{n}}{(n + 1)!}

f (x) = \sum_{n = 0}^{\infty} n! \frac{x^{n}}{n!}

f (x) = \sum_{n = 0}^{\infty} (n + 1)! \frac{x^{n}}{n!}

f (x) = \sum_{n = 0}^{\infty} (n + 1) x^{n}

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Verified step by step guidance

Step 1: Recall the definition of a Maclaurin series. A Maclaurin series is a special case of a Taylor series centered at x = 0. It is expressed as f(x) = ∑_{n=0}^{∞} f^{(n)}(0) * (x^n) / n!, where f^{(n)}(0) represents the nth derivative of f evaluated at x = 0.

Step 2: Substitute the given information f^{(n)}(0) = (n + 1)! into the Maclaurin series formula. This means the nth derivative of f at x = 0 is (n + 1)!. The series now becomes f(x) = ∑_{n=0}^{∞} (n + 1)! * (x^n) / n!.

Step 3: Simplify the expression. Notice that (n + 1)! can be rewritten as (n + 1) * n!. Substituting this into the series gives f(x) = ∑_{n=0}^{∞} [(n + 1) * n!] * (x^n) / n!.

Step 4: Cancel out n! in the numerator and denominator. This simplifies the series to f(x) = ∑_{n=0}^{∞} (n + 1) * x^n.

Step 5: Compare the simplified series f(x) = ∑_{n=0}^{∞} (n + 1) * x^n with the given options. The correct answer is f(x) = ∑_{n=0}^{∞} (n + 1) * x^n.