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sum_(m=0)^(oo)n!(cos ((1)/(sqrt(n!)))-1)

m=0n!(cos1n!1) \sum_{m=0}^{\infty} n !\left(\cos \frac{1}{\sqrt{n !}}-1\right)

Full solution

Q. m=0n!(cos1n!1) \sum_{m=0}^{\infty} n !\left(\cos \frac{1}{\sqrt{n !}}-1\right)
  1. Understand Behavior of Cosine Function: We are given the infinite series m=0n!(cos(1n!)1)\sum_{m=0}^{\infty}n!(\cos (\frac{1}{\sqrt{n!}})-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)\cos(x) around x=0x=0 is cos(x)=1x22!+x44!\cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \ldots . When xx is very small, cos(x)\cos(x) is approximately 1x221 - \frac{x^2}{2}. Since 1n!\frac{1}{\sqrt{n!}} becomes very small as nn increases, we can use this approximation for large nn.
  2. Analyzing Term for Large nn: Let's analyze the term cos(1/n!)1\cos(1/\sqrt{n!}) - 1 for large nn. Using the approximation cos(x)1x2/2\cos(x) \approx 1 - x^2/2 for small xx, we get cos(1/n!)1(1/n!)2/2=1/(2n!)\cos(1/\sqrt{n!}) - 1 \approx - (1/\sqrt{n!})^2/2 = -1/(2n!). This approximation becomes more accurate as nn increases.
  3. Multiplying by n!n!: Now, we multiply the approximated value by n!n! as per the series definition. This gives us n!×(1/(2n!))=1/2n! \times (-1/(2n!)) = -1/2 for large nn. However, we must be cautious because this approximation does not hold for small values of nn, specifically when nn is 00 or 11.
  4. Calculating for n=0n=0 and n=1n=1: For n=0n=0, the term is 0!0!(cos(1/0!)1\cos(1/\sqrt{0!}) - 1) = cos(1)1\cos(1) - 1, which is a finite number. For n=1n=1, the term is 1!1!(cos(1/1!)1\cos(1/\sqrt{1!}) - 1) = cos(1)1\cos(1) - 1, which is also a finite number. These terms do not follow the approximation and must be calculated separately.
  5. Explicit Calculation for First Two Terms: We calculate the first two terms explicitly. For n=0n=0, the term is cos(1)10.4597\cos(1) - 1 \approx -0.4597. For n=1n=1, the term is also cos(1)10.4597\cos(1) - 1 \approx -0.4597. So the sum of the first two terms is approximately 0.9194-0.9194.
  6. Using Approximation for n2n \geq 2: For n2n \geq 2, we can use the approximation 12-\frac{1}{2} for each term in the series. However, we must recognize that the series m=2(12)\sum_{m=2}^{\infty}(-\frac{1}{2}) does not converge, as it is the sum of an infinite number of constant terms. This means the series diverges to negative infinity.
  7. Recognizing Divergence: Since the series diverges for n2n \geq 2, the sum of the entire series from m=0m=0 to infinity also diverges. Therefore, the series does not have a finite sum.

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