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Here is the question:

enter image description here

Here is my work:

I substituted from n = 0 to n = 9 So, I got 1, -1, 0.5, -0.166, 0.0416, -0.0083, 0.00138, -0.000198, 0.0000248, -0.0000027. Then, I added all these numbers together and got $$\frac{16687}{45360}$$ Once, I got that number, I set up my interval, which was: $$\frac{16687}{45360} < S < \frac{16687}{45360} + 0.000005$$

That's my answer. Did I do it correctly or was I supposed to get an exact number?

amateurmath
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  • How do you know $S<{16687\over45360}+0.000005$? Where did you get that $0.000005$ from? – Gerry Myerson Mar 10 '22 at 06:38
  • @GerryMyerson I added the 5 x 10^-6 from the problem. I was following a khan academy video, but I probably misunderstood what he was saying. – amateurmath Mar 10 '22 at 06:40
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    The alternating series test implies that for any $N\geq 1$, $$ \sum\limits_{n = 0}^\infty {( - 1)^n \frac{1}{{n!}}} = \sum\limits_{n = 0}^{N - 1} {( - 1)^n \frac{1}{{n!}}} + \varepsilon _N , $$ where $$ \left| {\varepsilon _N } \right| \le \frac{1}{{N!}}. $$ You need to choose $N$ so that $\left| {\varepsilon _N } \right| \leq 5\cdot 10^{-6}$. The above estimate for $\left| {\varepsilon _N } \right| $ helps you to find such an $N$. – Gary Mar 10 '22 at 06:47
  • @Gary I'm not sure... I thought n = 9 was the maximum, but now I'm confused about how to solve this. Do I not use n = 10? – amateurmath Mar 10 '22 at 07:06
  • $N=9$ is the smallest value for which $\frac{1}{N!}\leq 5\cdot 10^{-6}$. So you can take $N=9$, i.e., the last term kept in the series is $\frac{1}{8!}$. Of course any larger $N$ would work as well, but you do not want to have too many terms. – Gary Mar 10 '22 at 07:07
  • Just by curiosity, have a look at the superb approximation @Gary gave in https://math.stackexchange.com/questions/430167/is-there-an-inverse-to-stirlings-approximation/461207#461207 – Claude Leibovici Mar 10 '22 at 07:13
  • @Gary So, rather than adding 5 * 10^-6, would I take the value of n = 9? Edit: Alright I'll check it out – amateurmath Mar 10 '22 at 07:13
  • The task is to figure out how many terms you have to keep so that their sum approximates the sum of the infinite series by an absolute error that is less than $5\cdot 10^{-6}$. The approximation is $$ 1 - 1 + \cdots + \frac{1}{{8!}} = \frac{{2119}}{{5760}}. $$I feel that you did not completely understand the various concepts here. – Gary Mar 10 '22 at 07:14
  • @Gary Yeah, I'm not really sure how to do this. We just started learning this week. I've been looking at different pdfs but they all seem confusing. How did you get 2119/5760? Did I get some parts right, like about adding the numbers I listed? – amateurmath Mar 10 '22 at 07:53
  • I think you added $-1/9!$ as well. That is why your rational approximation is different. But as I noted you do not have to go up to $-1/9!$ in order to get the required accuracy. – Gary Mar 10 '22 at 08:08

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