Calculus Help!!
Find the sum of the series
\[\sum_{n=0}^{\infty} 8^{n}/(9^{n}n!)\]

- anonymous

Calculus Help!!
Find the sum of the series
\[\sum_{n=0}^{\infty} 8^{n}/(9^{n}n!)\]

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- anonymous

i got 0 but it is incorrect!!

- perl

use ratio test?

- perl

first of all you have (8/9)^n

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## More answers

- anonymous

yes

- experimentX

it's a converging series ... but sum is obviously not zero.

- perl

cant be zero, plug in a few terms LOL

- anonymous

i got\[\lim_{n \rightarrow \infty}\] 8/9(n+1)

- perl

no thats just a condition to make sure it does not diverge

- anonymous

so it is not equal to sum?

- perl

no, the limit is just the limit of the sequence terms. not adding them

- anonymous

but how should i start with for adding them?

- 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

it diverges!!

- perl

does it? how do you know

- perl

no it doesnt diverge, that is a geometric series , it converges

- anonymous

i think the answer should be 9,am i right?

- anonymous

no!!sorry!!

- anonymous

is it should be a/(1-r)

- anonymous

The answer is e^(8/9)
e^x = Sum[ x^n/n! , {n,0, Infinity]]
x = 8/9

- anonymous

Thanks!!

- anonymous

but is it you can only get it when you knew some of the patterns of the functions?

- experimentX

somewhat seem to me like e times geometric progression of 8/9

- experimentX

since it is less than 1 ... it is definitely going to be less than e

- anonymous

it makes sense but in this case,we have to know e^x pattern in order to get this question?

- experimentX

we know that 1+1+1/2+1/3!+1/4! = e +1

- anonymous

okay!!

- experimentX

also we know (8/9)^t is a geometric progression

- anonymous

okay!!i think i got it!!Thanks everyone=))

- experimentX

still we haven't calculated the value yet ... we have just deduced that it's less than e

- anonymous

but is it you can only get it when you knew some of the patterns of the functions?
Yes

- 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

e^x = Sum[ x^n/n! , {n,0, Infinity]]

- experimentX

yeah ... he's right

- perl

ohh

- perl

elia, is there a test you can use to see if this converges

- perl

that was pretty cloever, i wouldnt have thought of that

- experimentX

tests are lot easier than fiding sum, use comparison test ... for best

- anonymous

@perl you can use the ratio test.

- perl

right, this was a stronger result

- experimentX

(8/9)^t/n! < 1/n!
since 1/n! converges, the LHS must also converge

- anonymous

you must know the patterns first!!

- perl

but then , how do you prove series 1/n! converges?

- perl

is that using ratio test?

- experimentX

it's a standard value 1/n! = e

- perl

true

- anonymous

@experimentX you still need the ratio test to show that 1/n! converges

- anonymous

it's a standard value 1/n! = e
Sum[ 1/$n!, {n,0, Infinity] =e

- experimentX

well, yes ... there it goes. 1/n!/1/(n+1)! -> 0 as n->inf

- perl

so in other words, you can prove that series 1/n! converges. it turns out that it equals e

- anonymous

If you use the ratio test with the original series, you do not need to do it again.

- experimentX

yeah that's right too ...

- 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

@perl yes.

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