Question : How do we prove that the sequence (1+(1/n))^n is bounded?

My thoughts on this question : We know that, to prove a sequence is bounded, we should prove that it is both bounded above and bounded below.

We can see that all the terms in the sequence is greater than 0. Which means the sequence is bounded below.
But gow do we prove that it is bounded above. Does taking the limit of the general term(say nth term) as n tends to infinity always gives us the highest possible term of the sequence, which will give us the opportunity to find an upper bound of the sequence?

Thanks!

Hmmm. By the Binomial Theorem,

(1 + 1/n)^n
= 1 + n*1/n + n(n-1)/2!n^2 + n(n-1)(n-2)/3!n^2 + ...
< 1 + 1 + 1/2! + 1/3! + ...
which you can show converges

I expect google can turn up various more direct proofs.

I Googled this and find that proof. But can you please let me know whether we can take the highest value of a monotonically increasing sequence by taking the limit of its general term(say nth term), when n goes to infinity?

Can we consider this value(if finite), as an upper bound of this sequence?

To prove that the sequence (1 + 1/n)^n is bounded, we need to show that it is both bounded above and bounded below.

First, let's consider the bounded below condition. We can see that all the terms in the sequence are greater than 0, as (1 + 1/n) is always positive. Thus, the sequence is bounded below by 0.

Now, let's move on to proving the bounded above condition. To do this, we will use a property of the sequence known as the Binomial Theorem. The Binomial Theorem states that for any real numbers a and b, and any positive integer n, the expansion of (a + b)^n can be written as:

(a + b)^n = C(n, 0)a^n + C(n, 1)a^(n-1)b + C(n, 2)a^(n-2)b^2 + ... + C(n, r)a^(n-r)b^r + ... + C(n, n)b^n,

where C(n, r) denotes the binomial coefficient, also known as "n choose r." The binomial coefficient C(n, r) is given by the formula:

C(n, r) = n! / (r!(n-r)!)

Now, let's take the sequence (1 + 1/n)^n and substitute it into the Binomial Theorem:

(1 + 1/n)^n = C(n, 0)(1^n) + C(n, 1)(1^(n-1))(1/n) + C(n, 2)(1^(n-2))(1/n)^2 + ... + C(n, r)(1^(n-r))(1/n)^r + ... + C(n, n)(1/n)^n

Note that in each term of the expansion, the power of 1 remains constant (as 1^n, 1^(n-1), 1^(n-2), etc., all equal 1). Additionally, the powers of 1/n decrease as we move through the terms.

Now, as n tends to infinity, each term with a positive power of 1/n approaches 0. This is because as n gets larger, 1/n becomes smaller and approaches 0. Thus, all terms in the expansion with a positive power of 1/n tend towards 0.

This leaves us with only two terms remaining in the expansion:

(1 + 1/n)^n = C(n, 0)(1^n) + C(n, n)(1/n)^n

Since C(n, 0) = 1 and C(n, n) = 1, we have:

(1 + 1/n)^n = 1 + (1/n)^n

Now, as n tends to infinity, the term (1/n)^n tends towards 0. This can be proven using limits and exponentiation properties. Therefore, the sequence (1 + 1/n)^n converges to a finite limit, which we denote as e (the base of the natural logarithm).

As a result, we can conclude that the sequence (1 + 1/n)^n is bounded above by e.

In summary, the sequence (1 + 1/n)^n is bounded both above (by e) and below (by 0), which proves its boundedness.

To prove that the sequence (1+(1/n))^n is bounded, we need to show that it is both bounded above and bounded below.

First, let's consider the lower bound. It is clear that all terms in the sequence are positive, as 1/n is positive for any positive integer n, and adding 1 to a positive number gives a positive result. Therefore, the sequence is bounded below by 0.

Next, let's focus on the upper bound. One common way to find the upper bound is by taking the limit of the sequence as n tends to infinity. If the limit exists and is finite, it can serve as an upper bound for the sequence.

Let's find the limit of the sequence (1+(1/n))^n as n tends to infinity:

lim(1+(1/n))^n as n approaches infinity

We can rewrite this as:

lim((1+(1/n))^(1/n))^n as n approaches infinity

Now, let's use the property of limits, where the limit of (a^b)^c is equal to a^(b * c):

lim((1+(1/n))^(1/n))^n as n approaches infinity
= lim((1+(1/n))^(1/n) * n) as n approaches infinity

Now, applying the limit property (lim a * b = lim a * lim b), we can separate the two limits:

lim(1+(1/n))^(1/n) * lim n as n approaches infinity
= (lim(1+(1/n))^(1/n)) * (lim n) as n approaches infinity

The limit of n as n approaches infinity is infinity. Now let's consider the limit of (1+(1/n))^(1/n) as n tends to infinity:

lim(1+(1/n))^(1/n) as n approaches infinity

This limit is a well-known limit that evaluates to e, where e is Euler's number (approximately 2.71828). So we have:

(e) * (infinity) as n approaches infinity

Since multiplying a finite number (e) by infinity results in infinity, we can conclude that the sequence (1+(1/n))^n is bounded above by infinity.

Combining this with the lower bound of 0, we can say that the sequence (1+(1/n))^n is bounded below by 0 and bounded above by infinity, which means it is bounded.