Question

perl's sum question:

Answer 1

\[\LARGE \sum_{n=1}^{2000} (\frac{1}{ \sum_{k=1}^n k!})\]

Answer 2

Is there a closed form for this?

Answer 3

right

Answer 4

I don't know, can you plug it into wolfram alpha like that?

Answer 5

i did before , but now i forgot the syntax

Answer 6

actually I used maple

Answer 7

let me screen shot it

Answer 8

I got 1.482622363

Answer 9

In simpler terms for other people who are curious, it looks like this: \[\Large \frac{1}{1!}+ \frac{1}{1!+2!}+ \frac{1}{1!+2!+3!}+...+ \frac{1}{1!+...+2000!}\]

Answer 10

right

Answer 11

Slightly larger than e/2

Answer 12

And of course, smaller than e.

Answer 13

first few sums produce the bulk of the sum, then it goes to zero fast

Answer 14

I wonder if in the limit as n goes to infinity if it approaches some sort of thing related to e?

Answer 15

if someone could put that into wolfram, that would be nice :)

Answer 16

is this considered a double series?

Answer 17

a double series is
Sum (Sum ( ... ) )

Answer 18

I don't think it is, at least not in this form it isn't.

Answer 19

wow sum (k!) is divisible by 3 . just saying

Answer 20

oh we might as well ask, does this converge as n goes to infinity?

Answer 21

I think so

Answer 22

Definitely, since this is the limit for e: \[\LARGE e= \sum_{n=0}^\infty \frac{1}{n!}\]

That's why I said it's less than e earlier.

Answer 23

yes it is , get fraction sum of subsequences

Answer 24

wait, its not in that form.
how did you see its slightly larger than e/2 , are you doing an inequality

Answer 25

I just calculated e/2 and compared it to your calculation. I knew it had to be less than e because the denominator is the same but contains more terms.

Answer 26

http://www.wolframalpha.com/input/?i=e%2F2%3C1.4826

Answer 27

1/sum(k!) < 1/2 (n)! for all n>2

Answer 28

so clear
sum (sum (k!) < 1/2 sum (1/n!) for n>2
sum(sum k!) <e/2

Answer 29

i dont know if this helps
http://www.wolframalpha.com/input/?i=1%2Fsum%28k!%2Ck%3D1%2Cn%29+

Answer 30

@Marki do you have a typo
sum (sum k! ) < e/2 ,
do you mean
sum ( 1/ sum k! ) < e2,
and i think that is false

e/2 < sum (1 / sum k! ) < e

Answer 31

@Marki do you have a typo
you wrote
sum (sum k! ) < e/2 ,
did you mean sum ( 1/ sum k! ) < e/2, and i think that is false

e/2 < sum (1 / sum k! ) < e

Answer 32

I don't think the fact that it is greater than e/2 is at all significant.

Answer 33

right, you can probably find a tighter bound

Answer 34

You might be able to play around this when n gets infinite they approach the same number. \[(1+\frac{1}{n})(1+\frac{1}{n})...(1+\frac{1}{n})=(1+\frac{1}{n})^n=\frac{1}{0!}+\frac{1}{1!}+...+\frac{1}{n!}\]

Answer 35

my original impulse was to try and turn the sum into a product , too

Answer 36

its all about comparison , as u can see that our sequence is less than 1/2 sum (1/n!)
the second one is converge thus the first one converge , however im trying to compute how to do it for infinite can't see anything with fourier xD if you have a magical function please show it :)

Answer 37

Another fun fact, the number is between cosh(1) and sinh(1). I say this because \[\large \cosh(1)= \frac{1}{0!}+ \frac{1}{2!}+ \frac{1}{4!}+...\]

Answer 38

and here is what i was trying to show u guys