Question

How to prove the limit as n tends to infinity of r^n/n! where r is some positive number.

Answer 1

Use the different tests from Calculus.

Answer 2

ratio test loves factorials i believe.

Answer 3

anytime you have a number raised to a power of n is an indication of the ratio test

Answer 4

I know that the limit is zero but how to prove it

Answer 5

is limit zero or inifinty?

Answer 6

zero as n tends to infinity

Answer 7

u want me to write out how to do the entire question?

Answer 8

\[\lim_{n \rightarrow \infty} r ^{n}/n!\]

Answer 9

Nope

Answer 10

Just how to do it

Answer 11

if u do the ratio test....you wind up with 1/n+1...and 1/infinity is 0

Answer 12

But does it means that it converges at 0

Answer 13

I thought ratio test are for series only??

Answer 14

it just means it converges....there's no sum of what it converges to..

Answer 15

This is not series

Answer 16

this is a sequence or a series?

Answer 17

if a series converges, the the sequence must also converge, and it has to converge to 0.

Answer 18

but the limit says to find the point of convergence

Answer 19

You cant have a series converge, and the sequence diverge, or converge to something other than 0.

Answer 20

it will tend to converge to 0 as that's the answer you have...but that's not "actually" what the sum of what it will converge to is

Answer 21

So how to get the limit?

Answer 22

when u get 1/n+1 you take the limit as n goes to infinity....that will give you 1/infinity which is 0

Answer 23

Joe, Divergence test?

Answer 24

does the question actually ask you for the sum of the series?

Answer 25

So you could solve this problem (in a roundabout way) by showing that the series:
\[\sum_{0}^{\infty}\frac{r^n}{n!}\]
converges....which it does. it converges to:
\[e^r\]
because the series converges, the sequence must converge to 0.

Answer 26

How about squeeze theorem or may be l' Hospital?

Answer 27

joemath is correct..but understand...0 is not the actual answer of what it converges to...it's simply what it tends to converge to

Answer 28

how will you take derivative of factorial? for l'hopital

Answer 29

or r^n?

Answer 30

Yeap, that's the problem so what about Squeeze theorem?

Answer 31

it's a ratio test....anytime u have something raised to a power of n...it's a ratio test

Answer 32

The ratio test would work as well (like sonofa is saying). it would show the series converges, and likewise the sequence converges to 0.

Answer 33

So according to ratio test if the value is 0 it converges and converges to 0 then....

Answer 34

the sequence converges to 0....you really can't determine the actual number of what the series will converge to...it just is convergent

Answer 35

if ratio is smaller than 1 converges
greater , diverge

Answer 36

inconclusive at 1

Answer 37

that series converges to e^x for sure. thats the power series for e^x (doesnt have to do with the problem, just saying :P lol)

Answer 38

yea...well he just asked for for the series...don't think we're testing endpoints here :)

Answer 39

What if the limit of ratio was 0.25 which means it converges but does it converges to 0.25 also

Answer 40

it's still just convergent

Answer 41

no, that just lets you know the series converges to something.

Answer 42

So, only if the limit of the ratio is 1, then I can safely say that it is convergent to 0 then

Answer 43

sorry 0

Answer 44

if u get 1...it's inconclusive

Answer 45

you show how you got 0 as an answer...and you just say it's convergent

Answer 46

it's just proof that you know what your doing and that your proving the series is convergent...but what it actually converges to is of no matter...don't think that deep into it...it's either convergent or divergent....if the question asks for the sum of what it converges to...you can trust that your in a geometric series where you can get an actual sum...outside of the geometric series...your not going to get an actual number of what it converges to....your just going to get a number...less than 1 is convergent, greater than 1 is divergent...the actual number doesn't matter all that much

Answer 47

Sorry again but how can u say that the limit is zero just by showing through ratio test that it is convergent

Answer 48

would it be easier if i typed it out in steps so u can visually see it?

Answer 49

I would love to

Answer 50

okay...gimme a few and let me type it out

Answer 51

Allrite, absolutely jaffa of a discussion this has been

Answer 52

\[series can be rewritten as \lim as n--> infinity of n/n!....step 1...ratio test. you get absolute value of (n+1)/(n+1)! / n/n! ...step 2...flip and multiply...you get absolute value of (n+1) n! / n (n+1)!....step 3...combine like terms...the n \in the numerator and denominator cancel leaving you a 1 \in the numerator and you can group the n! \in the numerator and denomator.....when you group n! \it goes \to 1 so basically goes away and your \left with 1/(n+1) which was \left \over. step 4. at this point you can take the limit as n goes \to infinity...if you plug infinity for n you get 1/infinity +1.....1/infinity is 0. Since 0 is less than one. We say the series is convergent. Does \not matter what \it actually converges \to. We got 0 for our limit, \therefore the series is convergent\]

Answer 53

Don't know why I can't see it properly though

Answer 54

yea i think it gotta retype it out

Answer 55

Thanks for ur effort

Answer 56

no problem..gimme a sec to retype

Answer 57

okay...so your series can be rewritten as the lim as n goes to infinity of the absolute value of the series. step 1. you get n+1/(n+1)! / n/n! ....step 2 you flip and multiply giving you n+1 n! in numerator and n (n+1)! in the denominator. step 3 you combine your terms. n's will cancel leaving you a 1 in the numerator. and the n!'s can be grouped...when their grouped they go to 1. so your left with 1/ n+1.....step 4...now you can take the limit. replace infinity with n and you get 1/infinity +1 which is still 1/infinity. 1/infinity is 0. Therefore the series converges according to the ratio test. it doesn't actually converge to zero, it's just a theorem. Without thinking too deep about it...just think surface. you got zero for an answer so the series converges...that's it. if you got 2, you'd say the series diverges. Those aren't actual numbers, just tests. so when you get 0 for the answer...you just say the series converges by the ratio test.

Answer 58

Thanks

Answer 59

yea no problem....i love helping..and i hope i'm making sense as to what im saying....for all we know this series can converge to sqrt of pi over pi which is some ridiculous number...but we got 0 for our test..so we know it converges and that's all we actually want...same as if the answer was 2...it could diverge to infinity a million times over for all we know...but we got 2 for our test..and we know it diverges..all that matters