Understand Behavior of Cosine Function: We are given the infinite series ∑m=0∞n!(cos(n!1)−1). To solve this, we need to understand the behavior of the cosine function as its argument approaches zero. The Taylor series expansion of cos(x) around x=0 is cos(x)=1−2!x2+4!x4−…. When x is very small, cos(x) is approximately 1−2x2. Since n!1 becomes very small as n increases, we can use this approximation for large n.
Analyzing Term for Large n: Let's analyze the term cos(1/n!)−1 for large n. Using the approximation cos(x)≈1−x2/2 for small x, we get cos(1/n!)−1≈−(1/n!)2/2=−1/(2n!). This approximation becomes more accurate as n increases.
Multiplying by n!: Now, we multiply the approximated value by n! as per the series definition. This gives us n!×(−1/(2n!))=−1/2 for large n. However, we must be cautious because this approximation does not hold for small values of n, specifically when n is 0 or 1.
Calculating for n=0 and n=1: For n=0, the term is 0!(cos(1/0!)−1) = cos(1)−1, which is a finite number. For n=1, the term is 1!(cos(1/1!)−1) = cos(1)−1, which is also a finite number. These terms do not follow the approximation and must be calculated separately.
Explicit Calculation for First Two Terms: We calculate the first two terms explicitly. For n=0, the term is cos(1)−1≈−0.4597. For n=1, the term is also cos(1)−1≈−0.4597. So the sum of the first two terms is approximately −0.9194.
Using Approximation for n≥2: For n≥2, we can use the approximation −21 for each term in the series. However, we must recognize that the series ∑m=2∞(−21) does not converge, as it is the sum of an infinite number of constant terms. This means the series diverges to negative infinity.
Recognizing Divergence: Since the series diverges for n≥2, the sum of the entire series from m=0 to infinity also diverges. Therefore, the series does not have a finite sum.
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