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Sagot :
Sure! To determine the convergence or divergence of the series [tex]\(\sum_{n=1}^{\infty} \frac{1}{n!}\)[/tex] using the Direct Comparison Test, let's follow these steps:
1. Identify the series to compare with:
Given the series [tex]\(\sum_{n=1}^{\infty} \frac{1}{n!}\)[/tex], we need to find a series that is known to converge or diverge, and that we can compare with [tex]\(\frac{1}{n!}\)[/tex].
2. Select a suitable comparison series:
Consider the geometric series [tex]\(\sum_{n=1}^{\infty} \frac{1}{2^n}\)[/tex]. This geometric series converges because its common ratio [tex]\(r = \frac{1}{2}\)[/tex] satisfies [tex]\(|r| < 1\)[/tex].
3. Show the inequality holds:
We need to show that [tex]\(\frac{1}{n!} \leq \frac{1}{2^n}\)[/tex] for sufficiently large [tex]\(n\)[/tex].
Let's examine the terms for different values of [tex]\(n\)[/tex]:
- For [tex]\(n = 1\)[/tex]: [tex]\(\frac{1}{1!} = 1 \quad \text{and} \quad \frac{1}{2^1} = \frac{1}{2}\)[/tex].
Here, [tex]\(\frac{1}{1!} \not\leq \frac{1}{2^1}\)[/tex].
- For [tex]\(n = 3\)[/tex]: [tex]\(\frac{1}{3!} = \frac{1}{6} \quad \text{and} \quad \frac{1}{2^3} = \frac{1}{8}\)[/tex].
Here, [tex]\(\frac{1}{3!} \leq \frac{1}{2^3}\)[/tex].
- For [tex]\(n = 4\)[/tex]: [tex]\(\frac{1}{4!} = \frac{1}{24} \quad \text{and} \quad \frac{1}{2^4} = \frac{1}{16}\)[/tex].
Here, [tex]\(\frac{1}{4!} \leq \frac{1}{2^4}\)[/tex].
4. Generalize for larger [tex]\(n\)[/tex]:
We observe that for [tex]\(n \geq 4\)[/tex], the factorial [tex]\(n!\)[/tex] grows much faster than the exponential [tex]\(2^n\)[/tex]. Thus, for sufficiently large [tex]\(n\)[/tex], the inequality [tex]\(\frac{1}{n!} \leq \frac{1}{2^n}\)[/tex] holds true.
5. Conclude by the Direct Comparison Test:
Since [tex]\(\sum_{n=1}^{\infty} \frac{1}{2^n}\)[/tex] converges and [tex]\(\frac{1}{n!} \leq \frac{1}{2^n}\)[/tex] for all [tex]\(n \geq 3\)[/tex], we conclude by the Direct Comparison Test that [tex]\(\sum_{n=3}^{\infty} \frac{1}{n!}\)[/tex] also converges.
Adding the finite terms [tex]\(\frac{1}{1!}\)[/tex] and [tex]\(\frac{1}{2!}\)[/tex] to a convergent series still results in a convergent series.
Therefore, we conclude that the series [tex]\(\sum_{n=1}^{\infty} \frac{1}{n!}\)[/tex] converges.
1. Identify the series to compare with:
Given the series [tex]\(\sum_{n=1}^{\infty} \frac{1}{n!}\)[/tex], we need to find a series that is known to converge or diverge, and that we can compare with [tex]\(\frac{1}{n!}\)[/tex].
2. Select a suitable comparison series:
Consider the geometric series [tex]\(\sum_{n=1}^{\infty} \frac{1}{2^n}\)[/tex]. This geometric series converges because its common ratio [tex]\(r = \frac{1}{2}\)[/tex] satisfies [tex]\(|r| < 1\)[/tex].
3. Show the inequality holds:
We need to show that [tex]\(\frac{1}{n!} \leq \frac{1}{2^n}\)[/tex] for sufficiently large [tex]\(n\)[/tex].
Let's examine the terms for different values of [tex]\(n\)[/tex]:
- For [tex]\(n = 1\)[/tex]: [tex]\(\frac{1}{1!} = 1 \quad \text{and} \quad \frac{1}{2^1} = \frac{1}{2}\)[/tex].
Here, [tex]\(\frac{1}{1!} \not\leq \frac{1}{2^1}\)[/tex].
- For [tex]\(n = 3\)[/tex]: [tex]\(\frac{1}{3!} = \frac{1}{6} \quad \text{and} \quad \frac{1}{2^3} = \frac{1}{8}\)[/tex].
Here, [tex]\(\frac{1}{3!} \leq \frac{1}{2^3}\)[/tex].
- For [tex]\(n = 4\)[/tex]: [tex]\(\frac{1}{4!} = \frac{1}{24} \quad \text{and} \quad \frac{1}{2^4} = \frac{1}{16}\)[/tex].
Here, [tex]\(\frac{1}{4!} \leq \frac{1}{2^4}\)[/tex].
4. Generalize for larger [tex]\(n\)[/tex]:
We observe that for [tex]\(n \geq 4\)[/tex], the factorial [tex]\(n!\)[/tex] grows much faster than the exponential [tex]\(2^n\)[/tex]. Thus, for sufficiently large [tex]\(n\)[/tex], the inequality [tex]\(\frac{1}{n!} \leq \frac{1}{2^n}\)[/tex] holds true.
5. Conclude by the Direct Comparison Test:
Since [tex]\(\sum_{n=1}^{\infty} \frac{1}{2^n}\)[/tex] converges and [tex]\(\frac{1}{n!} \leq \frac{1}{2^n}\)[/tex] for all [tex]\(n \geq 3\)[/tex], we conclude by the Direct Comparison Test that [tex]\(\sum_{n=3}^{\infty} \frac{1}{n!}\)[/tex] also converges.
Adding the finite terms [tex]\(\frac{1}{1!}\)[/tex] and [tex]\(\frac{1}{2!}\)[/tex] to a convergent series still results in a convergent series.
Therefore, we conclude that the series [tex]\(\sum_{n=1}^{\infty} \frac{1}{n!}\)[/tex] converges.
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