Unable to prove e without math. induction

[soopo]

Homework Statement

How can you prove the following statement without mathematical induction?
$$lim_{n -> \infty} (1 + \frac {1} {n})^{n} = \sum_{n=0}^{\infty} \frac {1} { n! }$$

The Attempt at a Solution

The statement can be proven by mathematical induction, but I am interested in how the sum statement can be deduced.

 

[Dick]

I'd be interested in how you prove that with induction. But try expanding (1+1/n)^n in a binomial expansion and look at the limit of the individual terms.

 

[soopo]

But try expanding (1+1/n)^n in a binomial expansion and look at the limit of the individual terms.

Let's consider only the first three and the last three terms in the binomial expansion
$$1 + (n,1) \frac {1} {n} + (n,2) \frac {1} {n^{2}} + (n,3) \frac {1}<br /> {n^{3}} + ... + (n, n-3) \frac {1} {n^{n-3}} + (n, n-2) \frac {1} {n^{n-2}} +<br /> (n, n-1) \frac {1} {n^{n-1}} + \frac {1} {n^{n}}$$

where (n,n) is a combination, for instance.
(n,0) and (n,n) are one.

Perhaps, I should now factorise by $$\frac {1} {n}$$.

 

[Dick]

Look at a single term, e.g. (n,3)/n^3. Expand (n,3) in factorials. Cancel some terms in the numerator and denominator. Can you see what happens?

 

[soopo]

Look at a single term, e.g. (n,3)/n^3. Expand (n,3) in factorials. Cancel some terms in the numerator and denominator. Can you see what happens?

$$1 + n (\frac {1}{n})^{1}) + n(n - 1) \frac {1}{2!} (\frac {1}{n})^{2}) + n(n - 1)(n - 2) \frac {1}{3!} (\frac {1}{n})^{3}) +<br /> ... + n(n - 1)(n - 2) \frac {1}{3!} (\frac {1}{n})^{n-3}) + n(n - 1) \frac {1}{2!} (\frac {1}{n})^{n-2}) + n (\frac {1}{n})^{n-1}) + \frac {1} {n^{n}}$$

Simplifying
$$2(1 + 1 + \frac {n - 1}{n} \frac {1}{2!} + (n - 1)(n - 2) \frac {1}{n^{2}*3!} +<br /> ... ) + \frac {1} {n^{n}}$$

The series diverges, since n is a positive integer.

We should somehow be able to get the sum notation.

 

[Dick]

You are forgetting the 1/n^k part. For the term containing 3! I get
(n)(n-1)(n-2)/(n^3*3!). What's the limit of that as n->infinity?

 

[soopo]

You are forgetting the 1/n^k part. For the term containing 3! I get
(n)(n-1)(n-2)/(n^3*3!). What's the limit of that as n->infinity?

I get
$$\frac {1 - 3/n - 2/n^2} {3!} => 1/3!,$$
as n goes to infinity.

We should now apply the result for the rest of the terms.

 

[Dick]

n/n->1. (n-1)/n->1. (n-2)/n->1. As n->infinity. I get that the limit is 1/3!. Your limit also goes to 1/3!. In spite of a sign error.

 

[soopo]

n/n->1. (n-1)/n->1. (n-2)/n->1. As n->infinity. I get that the limit is 1/3!. Your limit also goes to 1/3!. In spite of a sign error.

I agree with you.

I tried to prove the statement unsuccessfully by the proof of contradiction.
Nevertheless, if we can show that for n = 4 that the sum is 1/4!, we can apparently use mathematical induction to prove the statement.

I do not see any other way to prove the statement.

 

[Dick]

The sum isn't 1/3!. A single term converges to 1/3!. You may be confused because the n on the right side of the equation has nothing to do with the n on the left. Try proving lim n->infinity (1+1/n)^n=sum k=0 to infinity 1/k!.

 

[soopo]

The sum isn't 1/3!. A single term converges to 1/3!. You may be confused because the n on the right side of the equation has nothing to do with the n on the
left.

Thank you for the clarification!
I had an idea that there is a relation between n's at both sides.

Try proving lim n->infinity (1+1/n)^n=sum k=0 to infinity 1/k!.

The binomial expansion is finally
$$2 + \frac{n-1}{n*2!} + ... + \frac {1}{n^n}(\frac {n^3(n-1)} {2!} + n^2 + 1)$$

We know that each of the coefficients such as n/n -> 1, (n-1)/n -> 1 and (n-2)/n -> 1, as n goes to infinity.
In contrast, the other end of the sequence converges to zero.
1/n^n -> 0, n^(2-n) -> 0, and n^(3-n)(n-1)/2! -> 0,
as n goes to infinity.

We get the following sequence
$$1 + 1 + 1/2! + 1/3! + ...$$

Each term in the sequence is in line with the LHS of the equation.
This completes the proof.

---
Is the proof correct?

 

[sutupidmath]

Sorry for entering without permition, for i will not be that helpful either. But, without directly using induction, then it can be shown that:

$$\lim_{n\rightarrow \infty}\left(1+\frac{1}{n}\right)^n=e$$


And also, from the expansion of $$f(x)=e^x$$ using Taylor series around zero, and also letting x=1, we get


$$e=1+\frac{1}{1!}+\frac{1}{2!}+...=\sum_{n=0}^{\infty}\frac{1}{n!}$$

so from these two results we have that they are equal.

 

[Dick]

Thank you for the clarification!
I had an idea that there is a relation between n's at both sides.



The binomial expansion is finally
$$2 + \frac{n-1}{n*2!} + ... + \frac {1}{n^n}(\frac {n^3(n-1)} {2!} + n^2 + 1)$$

We know that each of the coefficients such as n/n -> 1, (n-1)/n -> 1 and (n-2)/n -> 1, as n goes to infinity.
In contrast, the other end of the sequence converges to zero.
1/n^n -> 0, n^(2-n) -> 0, and n^(3-n)(n-1)/2! -> 0,
as n goes to infinity.

We get the following sequence
$$1 + 1 + 1/2! + 1/3! + ...$$

Each term in the sequence is in line with the LHS of the equation.
This completes the proof.

---
Is the proof correct?

I think so. Each term in binomial expansion (n,k)*(1/n)^k converges to 1/k!. So it's plausible that the sum converges to the sum of 1/k! which is e. It's not horribly rigourous, but I've guessing that's what you are supposed to come up with.

 

[Dick]

Sorry for entering without permition, for i will not be that helpful either. But, without directly using induction, then it can be shown that:

$$\lim_{n\rightarrow \infty}\left(1+\frac{1}{n}\right)^n=e$$


And also, from the expansion of $$f(x)=e^x$$ using Taylor series around zero, and also letting x=1, we get


$$e=1+\frac{1}{1!}+\frac{1}{2!}+...=\sum_{n=0}^{\infty}\frac{1}{n!}$$

so from these two results we have that they are equal.

All true. But I think this 'proof' maybe supposed to bypass the knowledge that lim n->infinity (1+1/n)^n=e.

 

[soopo]

I think so. Each term in binomial expansion (n,k)*(1/n)^k converges to 1/k!. So it's plausible that the sum converges to the sum of 1/k! which is e. It's not horribly rigourous, but I've guessing that's what you are supposed to come up with.

How would you do the proof more rigorous?

 

[Dick]

How would you do the proof more rigorous?

You'd have to come with with some kind of error estimate for the sum of all of the parts that still contain n. But I wouldn't bother, because as stupidmath said, you know the limit is e and you know that the sum of 1/k! is also a convergent series summing to e from the Taylor series expansion of e^x.

 

[JJBladester]

How would you do the proof more rigorous?

Here is your rigorous proof. Using L'Hôpital's Rule,

$$Let y = lim_{n\rightarrow\infty}\left(1+\frac{1}{n}\right)^{n}$$

Take the natural logarithm of both sides of this equation.

$$ln(y) = ln\left[lim_{n\rightarrow\infty}\left(1+\frac{1}{n}\right)^{n}\right]$$

$$= lim_{n\rightarrow\infty}\left[ln\left(1+\frac{1}{n}\right)^{n}\right]$$

$$= lim_{n\rightarrow\infty}\left[nln\left(1+\frac{1}{n}\right)\right]$$

$$= lim_{n\rightarrow\infty}\left[\frac{ln \left(1+\frac{1}{n}\right)}{\frac{1}{n}}\right]$$

If we try to solve the above limit using direct substitution, we'll obtain the indeterminant form $$0/0$$. Thus, we apply L'Hôpital's Rule to attempt to find the limit by taking the derivative of the numerator and denominator and finding the limit of that ratio.

$$ln(y) = lim_{n\rightarrow\infty}\left[ln\left(\frac{1+\frac{1}{n}}{\frac{1}{n}}\right)\right]$$

$$= lim_{n\rightarrow\infty}\left[\left(\frac{-1/n^{2}}{1+\frac{1}{n}}\right)/\left(-1/n^2\right)}\right]$$

$$= lim_{n\rightarrow\infty}\left[1+\frac{1}{n}\right]$$
$$= 1$$

Therefore $$ln(y) = 1$$ which implies y = e by definition of logarithms.

By comparison, since we let y = the original limit expression, then:

$$e = lim_{n\rightarrow\infty}\left(1+\frac{1}{n}\right)^{n}$$

This completes the proof and shows why y converges to e as n gets larger and larger (approaches $$\infty$$).

 

[soopo]

$$ln(y) = lim_{n\rightarrow\infty}\left[ln\left(\frac{1+\frac{1}{n}}{\frac{1}{n}}\right)\right]$$

$$= lim_{n\rightarrow\infty}\left[\left(\frac{-1/n^{2}}{1+\frac{1}{n}}\right)/\left(-1/n^2\right)}\right]$$

$$= lim_{n\rightarrow\infty}\left[1+\frac{1}{n}\right]$$
$$= 1$$

There is perhaps something wrong in the statements following the expression

$$ln(y) = lim_{n\rightarrow\infty}\left[ln\left(\frac{1+\frac{1}{n}}{\frac{1}{n}}\right)\right]$$

The problem seems to be in the derivate.
I get
$$ln(y) = lim_{n\rightarrow\infty}\left[\left(\frac {1/n} {1 + \frac{1}{n}})\right]$$