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# Convergent Sequence watch

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1. "Let a be a real number. Show that (a^n)/n! converges to 0".

This looks very simple and I'm fairly sure it can be done just from the definition of convergence (the "epsilon" one, as opposed to "it gets very small"), but I think I'm missing something simple.

Obviously the case where |a|<1 is easy, but for the rest I'm not sure how to get rid of the a^n.

Thanks for any help.
2. Show that the sequence is strictly decreasing. Then show that for large enough n, a^n/n! < something which tends to 0.
3. |a^n/n!| is decreasing for all sufficiently large n. [You don't need to say "strictly" but you can if you like.] Since the sequence (|a^n/n!|) is bounded below by 0, it converges. Let L be the limit:

|a^n/n!| -> L as n -> infinity . . . . . (*)

Then

[|a|/(n + 1)] |a^n/n!| = | a^(n + 1)/(n + 1)! | . . . . . . (**)

and, using (*) twice,

LHS of (**) -> 0 * L = 0 as n -> infinity
RHS of (**) -> L as n -> infinity

So L = 0,

|a^n/n!| -> 0 as n -> infinity
a^n/n! -> 0 as n -> infinity

--

Alternatively, you can view the result as a consequence of the convergence of the series exp(a) = (sum for n = 0 to infinity) a^n/n!.
4. Thanks for the ideas. They seem to be fine. I was expecting methods more like starting witth |a^n/n! - 0| and showing it is less than or equal to epsilon for all epsilon >0, but I was maybe being too closed minded!
5. (Original post by sat)
"Let a be a real number. Show that (a^n)/n! converges to 0".

This looks very simple and I'm fairly sure it can be done just from the definition of convergence (the "epsilon" one, as opposed to "it gets very small"), but I think I'm missing something simple.

Obviously the case where |a|<1 is easy, but for the rest I'm not sure how to get rid of the a^n.

Thanks for any help.
can you not just say:

for any real a, there is always an integer b (say) where b is the lowest integer such that b>a. and the same for b, with c, and the same for c. there are infinitely many integers above a.

as n tends to infinity, more of these integers are being included in the multiplication which makes up the denominator

then the logic follows:

- as n tends to infinity, n! will eventually have a factor b (as defined above)
- once this point is passed, the value of a^n/n! will change such that:
(a) the numerator is multipled by a
(b) the denominator is multipled by an integer x such that x>a

since you have done the case for a<1, x>a>1 and so the denominator will increase more than the numerator. repeat to infinity and the ratio will tend to zero..

not quite rigourous, but perhaps with more numbers and less words is the theory right>
6. ok try again:
prove that as n tends to infinity, a^n/n! tends to zero, for a>1.

a is a real number above zero. a+1 is a real number 1 larger than a. define integer x as: a < x =< a+1
as n tends to infinity, (n is an integer I hope), there will be a point where n=x.

at every step from a^n/n! to a^(n+1)/(n+1)!, the step is effectively going from:
a*a^n/n!(n+1) = (a^n/n!)[a/(n+1)]

now for a > (n+1), this step will increase the expression, as a/(n+1) > 1. But as n tends to infinity, past the point to where a < (n+1), the expression before the step shown above is larger than the expression after, since a/(n+1) < 1. therefore the expression is decreasing - after this point the expression will be continually be multipled by a number (a/(n+1)) which itself tends to zero as n tends to infinity and a remains a constant real. as this is continued to infinity, but the factor (a/(n+1)) never falls below zero [as neither a nor n is negative].

erm.. yeah. I kinda fell into the realm of wordyness again.
7. (Original post by sat)
Thanks for the ideas. They seem to be fine. I was expecting methods more like starting witth |a^n/n! - 0| and showing it is less than or equal to epsilon for all epsilon >0, but I was maybe being too closed minded!
You asked for it! [Credit to mik1a - the idea of my proof is in his post. But I have succeeded in making the answer look incredibly complicated.]

Assume a is not zero.

Let epsilon > 0.

Define

M
= ceiling(|a|)
= smallest integer that is at least |a|

K = |a|^M/M! > 0

N = M + max{ 1, ceiling[ln(epsilon/K) / ln(M/(M + 1))] }

Then for all n >= N we have n >= M + 1 and

|a|^n/n!
= [|a|^M/M!] (product from i = 1 to n - M) |a|/(M + i)
= K (product from i = 1 to n - M) |a|/(M + i)
<= K (product from i = 1 to n - M) M/(M + i)
<= K (product from i = 1 to n - M) M/(M + 1)
= K [M/(M + 1)]^(n - M)
<= K [M/(M + 1)]^[ln(epsilon/K) / ln(M/(M + 1))]
= K * epsilon/K
= epsilon
8. (Original post by Jonny W)

Assume a is not zero.

Let epsilon > 0.

Define

M
= ceiling(|a|)
= smallest integer that is at least |a|

K = |a|^M/M! > 0

N = M + max{ 1, ceiling[ln(epsilon/K) / ln(M/(M + 1))] }

Then for all n >= N we have n >= M + 1 and

|a|^n/n!
= [|a|^M/M!] (product from i = 1 to n - M) |a|/(M + i)
= K (product from i = 1 to n - M) |a|/(M + i)
<= K (product from i = 1 to n - M) M/(M + 1)
= K [M/(M + 1)]^(n - M)
<= K [M/(M + 1)]^[ln(epsilon/K) / ln(M/(M + 1))]
= K * epsilon/K
= epsilon
Is this analysis?
9. (Original post by Euclid)
Is this analysis?
Anything that starts "let epsilon > 0" is analysis.
10. (Original post by Jonny W)
Anything that starts "let epsilon > 0" is analysis.
Good stuff! Im starting my very first Analysis course tomorrow !
11. A person who can, within a year, solve x² - 92y² = 1 is a mathematician

how the hell... I've been trying this all afternoon rearranging stuff and stuff. can't see any kind of factorisation
and plotting x against y looke nightmarish... is it a circle? no it cant be...

argh!
12. (Original post by mik1a)
A person who can, within a year, solve x² - 92y² = 1 is a mathematician

how the hell... I've been trying this all afternoon rearranging stuff and stuff. can't see any kind of factorisation
and plotting x against y looke nightmarish... is it a circle? no it cant be...

argh!
Precisely! It's very difficult, hence within a year. I haven't actually tried it myself, but I might do so now that it has caught attention
13. It's definately not a circle. At first I thought it was a hyperbola but I keep getting two different hyperbola's for the same equation - that's rather odd

I think the question is supposed to be aiming at finding non-zero integer solutions, because by inspection two solutions are x=1 y=0 and x=-1 y=0, hence making the question sound stupid.

Euclid
14. (Original post by Euclid)
It's definately not a circle. At first I thought it was a hyperbola but I keep getting two different hyperbola's for the same equation - that's rather odd

I think the question is supposed to be aiming at finding non-zero integer solutions, because by inspection two solutions are x=1 y=0 and x=-1 y=0, hence making the question sound stupid.

Euclid
Its a Pell equation, there's lots of information on methods to solve it on the net.

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