the proof of limit as n goes to infinity of (1+1/n)^n = e ???

Question

how can i prooving the limit as n goes to infinity of (1+1/n)^n = e?

Answer 1

just another math guy may be correct, but there is a much simpler solution using l'Hospital's rule.
(In this proof, all limits are as n→∞)
f(n) = (1+1/n)^n = [(n+1)/n] ^ n
= (n+1)^n / n^n
Take the derivitive of numerator and denomerator to find an equivalent limit.
lim f(n) = lim [(n+1)^n * (n/(n+1) + ln(n+1))] / [n^n * (1+ln(n))]
= lim (n+1)^n / n^n * { [n/(n+1) + ln(n+1)] / [1+ln(n) ] }
= lim f(n) * { [n/(n+1) + ln(n+1)] / [1+ln(n) ] }

Therefore, lim { [n/(n+1) + ln(n+1)] / [1+ln(n) ] } = 1 !
or lim n/(n+1) + ln(n+1) = lim (1 + ln(n) )
or, in the limit,
ln(n+1) - ln(n) = 1/(n+1)
ln[(n+1)/n] = 1/(n+1)
e^(1/(n+1)) = (n+1)/n = 1+1/n
or e = lim (1+1/n)^(n+1) = lim (1+1/n)^n

Answer 2

The first poster made no explicit use of the number e. 2^h-1 is small (so "approximately" equal to h), and 1000000000000000^h-1 is small too, for small enough h. It is true that e^h-1 is approximately equal to h in a well-defined way, where 2^h-1 and the other example I just said above are not, but this uses some extra properties of the number e.

This limit is often taken as the definition of e. One shows that the sequence ((1+1/n)^n) is increasing and bounded above so that it converges, and then we call the limit e. The fact that this limit converges at all is another interesting proof with various "tricks" that people use to show increasing and bounded above.

So your answer depends on how you define e, if it is not by using this limit. If, instead, you define e to be the sum of 1/n! as n goes from 0 to infinity, then you have something to prove. First note that the sum I mentioned actually converges. You can use the ratio test to see this, or you can estimate it directly to see that the sequence of partial sums is increasing and bounded above by 3.

Use the binomial theorem to see that
(1+1/n)^n = 1 + (n/1!)(1/n) + (n(n+1)/2!)(1/n)^2 + ... + (n!/((n-k)!k!))(1/n)^k + ... + (n!/n!)((1/n)^n
= 1 + (1/1!)(1/n) + (1/2!)(1-0/n)(1-1/n) + ... + (1/k!) (1-1/n)(1-2/n) ... (1-(k-1)/n) + ... + (1/n!)(1-0/n)(1-1/n)(1-2/n) ... (1-(n-1)/n)
<=1 + 1/1! + 1/2! + 1/3! + ... + 1/n!

Taking the limsup of both sides shows that the limsup of (1+1/n)^n is less than or equal to e under the sum definition of e. Recall that the limsup is equal to the limit when the limit exists.

Now we'll show that e is less than or equal to the lininf of (1+1/n)^n. Fix a positive interger k and consider the following sum for n>k.
1 + (1/1!)(1/n) + (1/2!)(1-1/n) + ... + (1/k!) (1-1/n)(1-2/n) ... (1-(k-1)/n).
As was shown above, this is smaller than (1+1/n)^n. Hence, taking the liminf as n goes to infinity, you get that 1 + 1/1! + 1/2! + ... + 1/k! <= liminf (1+1/n)^n. Well, now take the limit of *this* inequality as k goes to infinity. This shows that e is less than or equal to liminf (1+1/n)^n.

We've shown that limsup (1+1/n)^n <= e <= liminf (1+1/n)^n. It is a general fact that the liminf of a sequence is less than or equal to the limsup of a sequence. Hence, limsup (1+1/n)^n <= e <= liminf (1+1/n)^n <= limsup (1+1/n)^n. Hence, all of these terms are equal. In particular, your limit exists and equals e.

There are many other possible definitions of e, so if the above sum is not how you have defined e originally and the limit you mention is not how you defined e originally, you'll have to state your definition to get an appropriate answer.

Answer 3

e^h -1 for very small values of h is approximately = h
So e^h is approximately = h +1
So e is approximately = (h+1)^(1/h)
So lim (1+h)^(1/h) =e
h-> 0
Or, set t=1/h, then
lim (1+1/t)^t =e
t-> infinity

For more formal proofs see Fine, Calculus, page 71 or Smail, Elements of the Theory of infinite processes, pages 38-41.