The question: for $n=1$:$u_n=0.1$ Help
Test the following series for convergence or divergence:
$$
\frac{1!}{10}-\frac{2!}{10^2}+\frac{3!}{10^3}-\frac{4!}{10^4}+\cdots
$$
My answer:
The general term is then
$$
\sum_{n=1}^\infty\frac{(-1)^{n-1}n!}{10^n}
$$
and using the alternating series test $u_{n+1}
for $n=2$:$u_n=0.02$
for $n=3$:$u_n=0.006$
for $n=4$:$u_n=0.0024$
$\cdots$
Why does the series $\sum_{n=1}^\infty\frac{(-1)^{n-1}n!}{10^n}$ not converge?
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2Hint: For $n\geq 10$, $u_n\geq u_{n-1}$ – 2012-11-20
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0The series $\sum \frac{1}{u_n}$ converges, so... – 2012-11-24
3 Answers
Short answer: $\lim_{n\to\infty}\dfrac{n!}{10^n}=\infty$, so that $\dfrac{(-1)^{n-1}n!}{10^n}$ does nort converge to $0$.
The general term does not go to 0. Therefore, the sum does not converge. You can not determine if the terms $u_n$ go to 0 just by looking at a few of them, i.e., you can not determine a limit by looking at a few values. When $n$ is small, increasing by 1 makes the denominator increase by a factor of 10, but the numerator only increases by a factor of 1 or 2 or 3. But, once $n \geq 10$, now the numerator starts increasing faster than the denominator.
In this case, $\lim_{n\to \infty} u_n = \infty$. Do you see why?
Consider $|\frac{u_{n+1}}{u_n}|=\frac{n+1}{10}$. $\lim_{n \to \infty}|\frac{u_{n+1}}{u_n}| > 1$. So the series diverges, by ratio test.
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0I don't like the choise of words here, because convergence to $1$ doesn't mean anything about the convergence of the series. – 2012-11-20
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0@Belgi: It is a sufficient condition. $L\to \infty$ means the series diverges. – 2012-11-20
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0But thats not written, you gave the argument "does not converge to 1" which says nothing...still, "This diverges away from 1 " doesn't say what you want - it can diverge from $1$ when it convergers to say $0.5$. just write L tends to infinity as in the comment or say that the elements of the sequence does not tend to $0$... – 2012-11-20
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0@Belgi: ya thanks, got to have been careful about these things! – 2012-11-20
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0I have changed my vote accordingly :) – 2012-11-20