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

Question

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

anonymous · Answer

i got 0 but it is incorrect!!

perl · Answer

use ratio test?

perl · Answer

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

anonymous · Answer

yes

experimentx · Answer

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

perl · Answer

cant be zero, plug in a few terms LOL

anonymous · Answer

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

perl · Answer

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

anonymous · Answer

so it is not equal to sum?

perl · Answer

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

anonymous · Answer

but how should i start with for adding them?

perl · Answer

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 · Answer

it diverges!!

perl · Answer

does it? how do you know

perl · Answer

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

anonymous · Answer

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

anonymous · Answer

no!!sorry!!

anonymous · Answer

is it should be a/(1-r)

anonymous · Answer

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

anonymous · Answer

Thanks!!

anonymous · Answer

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

experimentx · Answer

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

experimentx · Answer

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

anonymous · Answer

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

experimentx · Answer

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

anonymous · Answer

okay!!

experimentx · Answer

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

anonymous · Answer

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

experimentx · Answer

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

anonymous · Answer

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

anonymous · Answer

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 · Answer

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

experimentx · Answer

yeah ... he's right

perl · Answer

ohh

perl · Answer

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

perl · Answer

that was pretty cloever, i wouldnt have thought of that

experimentx · Answer

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

anonymous · Answer

@perl you can use the ratio test.

perl · Answer

right, this was a stronger result

experimentx · Answer

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

anonymous · Answer

you must know the patterns first!!

perl · Answer

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

perl · Answer

is that using ratio test?

experimentx · Answer

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

perl · Answer

true

anonymous · Answer

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

anonymous · Answer

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

experimentx · Answer

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

perl · Answer

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

anonymous · Answer

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

experimentx · Answer

yeah that's right too ...

perl · Answer

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 · Answer

@perl yes.