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MathGroup Archive 2006

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Re: Re: Problem with limiits


On 18 Apr 2006, at 19:56, Roger Bagula wrote:

> Andrzej Kozlowski wrote:
>
>>
>> For any sequence of real numbers a1,a2, a3 .... converging to a real
>> number L,  any infinite subsequence of it will converge to the same
>> limit L. This is a very basic fact and is in fact true in any
>> complete metric space (a subsequence of a convergent sequence is a
>> Cauchy sequence, and therefore, in a complete metric space, is itself
>> convergent.)
>>
>> Given the above,  you only need the fact that there are infinitely
>> many primes (proved by Euclid and before him by Eudoxus).
>>
>> Andrzej Kozlowski
>>
> What you are saying is that if I have a composite number c[m]
> them the Limit:
>
> Limit[(1+1/c[m])^c[m],m->Infinity]=E
>
> as well.
> What About:
> Limit[(1-1/m)^m,m->Infinity] =1/E
>
> Your reasoning gives:
> Limit[(1-1/c[n])^c[n],n->Infinity]*Limit[(1+1/Prime[n])^Prime[n],N- 
> >Infinity]=1
> and
> Limit[(1-1/m)^m,m->Infinity]*Limit[(1+1/m)^m,m->Infinity]=1
> This does give:
> Limit[(1 - 1/m^2)^m, m -> Infinity]=1
>
> Thanks for your help.
>

No, I am saying that you have to have a *sub-sequence*.

(1+1/Prime[n])^Prime[n] for n from 1 to Infinity  is a sub-sequence  
of (1 + 1/m)^m from 1 to Infinity. If you can't see this general fact  
just look at


Table[(1 + 1/Prime[n])^Prime[n], {n, 1, 3}]


{9/4, 64/27, 7776/3125}

and


Table[(1 + 1/m)^m, {m, 1, 5}]


{2, 9/4, 64/27, 625/256, 7776/3125}

and it should become clear.

On the other hand:

(1-1/m)^m for m from 1 to Infinity is not a sub-sequence of (1 + 1/n) 
^n from 1 to Infinity.

Just look at the first few terms:


Table[(1 - 1/m)^m, {m, 1, 5}]


{0, 1/4, 8/27, 81/256, 1024/3125}


Do I need to say more?

Andrzej Kozlowski



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