anonymous
  • anonymous
Calculus Help!! Find the sum of the series \[\sum_{n=0}^{\infty} 8^{n}/(9^{n}n!)\]
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
i got 0 but it is incorrect!!
perl
  • perl
use ratio test?
perl
  • perl
first of all you have (8/9)^n

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More answers

anonymous
  • anonymous
yes
experimentX
  • experimentX
it's a converging series ... but sum is obviously not zero.
perl
  • perl
cant be zero, plug in a few terms LOL
anonymous
  • anonymous
i got\[\lim_{n \rightarrow \infty}\] 8/9(n+1)
perl
  • perl
no thats just a condition to make sure it does not diverge
anonymous
  • anonymous
so it is not equal to sum?
perl
  • perl
no, the limit is just the limit of the sequence terms. not adding them
anonymous
  • anonymous
but how should i start with for adding them?
perl
  • perl
suppose your sequence is {1/2^n} , n=1,2,3... so your sequence is 1/2, 1/4, 1/8, 1/16, ... lim (1/2^n) =0, does the series (the sum of the sequence terms) equal zero? of course not!!! 1/2 + 1/4 + 1/8 + ... clearly is not zero the question is, does it diverge ?
anonymous
  • anonymous
it diverges!!
perl
  • perl
does it? how do you know
perl
  • perl
no it doesnt diverge, that is a geometric series , it converges
anonymous
  • anonymous
i think the answer should be 9,am i right?
anonymous
  • anonymous
no!!sorry!!
anonymous
  • anonymous
is it should be a/(1-r)
anonymous
  • anonymous
The answer is e^(8/9) e^x = Sum[ x^n/n! , {n,0, Infinity]] x = 8/9
anonymous
  • anonymous
Thanks!!
anonymous
  • anonymous
but is it you can only get it when you knew some of the patterns of the functions?
experimentX
  • experimentX
somewhat seem to me like e times geometric progression of 8/9
experimentX
  • experimentX
since it is less than 1 ... it is definitely going to be less than e
anonymous
  • anonymous
it makes sense but in this case,we have to know e^x pattern in order to get this question?
experimentX
  • experimentX
we know that 1+1+1/2+1/3!+1/4! = e +1
anonymous
  • anonymous
okay!!
experimentX
  • experimentX
also we know (8/9)^t is a geometric progression
anonymous
  • anonymous
okay!!i think i got it!!Thanks everyone=))
experimentX
  • experimentX
still we haven't calculated the value yet ... we have just deduced that it's less than e
anonymous
  • anonymous
but is it you can only get it when you knew some of the patterns of the functions? Yes
anonymous
  • anonymous
i put e^(8/9)...according to @eliassaab and it makes sense now.since the sum of the e^x is e^x/n!
anonymous
  • anonymous
e^x = Sum[ x^n/n! , {n,0, Infinity]]
experimentX
  • experimentX
yeah ... he's right
perl
  • perl
ohh
perl
  • perl
elia, is there a test you can use to see if this converges
perl
  • perl
that was pretty cloever, i wouldnt have thought of that
experimentX
  • experimentX
tests are lot easier than fiding sum, use comparison test ... for best
anonymous
  • anonymous
@perl you can use the ratio test.
perl
  • perl
right, this was a stronger result
experimentX
  • experimentX
(8/9)^t/n! < 1/n! since 1/n! converges, the LHS must also converge
anonymous
  • anonymous
you must know the patterns first!!
perl
  • perl
but then , how do you prove series 1/n! converges?
perl
  • perl
is that using ratio test?
experimentX
  • experimentX
it's a standard value 1/n! = e
perl
  • perl
true
anonymous
  • anonymous
@experimentX you still need the ratio test to show that 1/n! converges
anonymous
  • anonymous
it's a standard value 1/n! = e Sum[ 1/$n!, {n,0, Infinity] =e
experimentX
  • experimentX
well, yes ... there it goes. 1/n!/1/(n+1)! -> 0 as n->inf
perl
  • perl
so in other words, you can prove that series 1/n! converges. it turns out that it equals e
anonymous
  • anonymous
If you use the ratio test with the original series, you do not need to do it again.
experimentX
  • experimentX
yeah that's right too ...
perl
  • perl
eliassab, i wanted to ask you about last question. it is sufficient to know that limit of f(x) exists at x =0. you dont need continuity there
anonymous
  • anonymous
@perl yes.

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