Finding the sum of following series $$\frac{1}{1!}+\frac{1}{2!} +...........+\frac{1}{n!}$$ I tried $$\frac{1}{n!}\Bigg[\frac{n!}{1}+\frac{n!}{2!}+\frac{n!}{3!}+..... .....+1\Bigg]$$ And end up getting $$\frac{1}{n!}\Bigg[(n-1)!\binom{n}{1}+(n-2)!\binom{n}{2}+........+\binom{n}{n}\Bigg]$$ After which I am stuck, But after this I am not able to solve it And hence i further want to find $$\sum_{n=3}^{2016} \frac{1}{n!}\Bigg[1-\frac{1}{n}\Bigg]$$ by using the result But I am not able to reach anywhere so far. It goes to a finite number
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https://math.stackexchange.com/questions/227551/sum-k-1-2-3-cdots-n-is-there-a-generic-formula-for-this your problem is the same but divided by 2016! – NewGuy Jan 28 '18 at 08:14
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@NewGuy Not at all. That question deals with sums of factorials, this deals with sums of their reciprocals. There's no division by 2016 -- I don't see where you're getting that. – saulspatz Jan 28 '18 at 08:17
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Are you looking for $\frac{1}{1!}+\frac{1}{2!} +...........+\frac{1}{n!}$? – user Jan 28 '18 at 08:17
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yes, and then the next one – Rishi Kakkar Jan 28 '18 at 08:18
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Do you know about Taylor series? – saulspatz Jan 28 '18 at 08:19
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A little idea but I am pretty sure it is not of those – Rishi Kakkar Jan 28 '18 at 08:19
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Possible duplicate of Finding the sum of this alternating series with factorial denominator. – Guy Fsone Jan 28 '18 at 08:21
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Please at least look at the question – Rishi Kakkar Jan 28 '18 at 08:21
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It's not a Taylor series, you can use Taylor series to get the value. My point is that there is not reason to sum the terms out to $n=2016$. $2016!$ is so big that it's reciprocal is effectively $0$. You will not be able to get a closed form expression for this sum. What you want to do is to calculate enough terms to approximate the sum to the desired accuracy. To do that, you have to estimate the sum of the terms you're leaving out. Taylor series would be useful for that, though there are other ways. – saulspatz Jan 28 '18 at 08:24
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As I said, you won't be able to find a closed-form expression for these sums. Let's take $\sum_{n=0}^{2016}{\frac{1}{n!}}$ Factorials get big rather quickly. Let's say you compute the sum out to $\frac{1}{20!}$ How much are you leaving out? The next term is $\frac{1}{21!} = \frac{1}{21}\frac{1}{20!}<\frac{1}{20}\frac{1}{20!}$. The term after that is $\frac{1}{22!} = \frac{1}{21\cdot 22}\frac{1}{20!}<\frac{1}{20^2}\frac{1}{20!},$ and so on. The total error is less than $$\frac{1}{20!}\sum_{n=1}^{\infty}{\frac{1}{20^n}} = \frac{19}{20!} \approx 2.163325 \cdot 10^{-20}$$ Even computing 20 terms is probably overkill.
You can make the same sort of analysis for the other series. Add them up though 20 terms and forget it.
saulspatz
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