Final Answer:
The sequence $\left(1 + \frac{1}{n}\right)^n$ is monotone increasing and bounded below by 2 and above by 3.
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Explanation:
Let’s break this down step by step so it makes sense.
We’re looking at the sequence where each term is $\left(1 + \frac{1}{n}\right)^n$, and $n$ starts at 1 and goes up: 1, 2, 3, 4, ...
Step 1: Is it increasing?
That means each term should be bigger than the one before.
For example:
- When $n = 1$: $\left(1 + \frac{1}{1}\right)^1 = 2$
- When $n = 2$: $\left(1 + \frac{1}{2}\right)^2 = (1.5)^2 = 2.25$
- When $n = 3$: $\left(1 + \frac{1}{3}\right)^3 ≈ (1.333)^3 ≈ 2.37$
You can see it’s going up! And it keeps going up as $n$ gets bigger — that’s what “monotone increasing” means.
Step 2: Is it bounded below by 2?
The first term is 2, and since it’s increasing, every next term is bigger than 2. So yes — it never goes below 2.
Step 3: Is it bounded above by 3?
This is trickier, but we can use a hint from the worksheet. It says:
> From this one can conclude that the sequence is increasing and that
> $2 \leq \left(1 + \frac{1}{n}\right)^n \leq 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \cdots + \frac{1}{n!} < 3$
That long sum on the right is like adding fractions with factorials in the bottom:
$1 + 1 + \frac{1}{2} + \frac{1}{6} + \frac{1}{24} + \cdots$
Even if you add them all up forever, it never reaches 3 — it stops around 2.718... (which is Euler’s number, e).
So since our sequence is less than that sum, and that sum is less than 3, our sequence is also less than 3.
✔ So, the sequence is:
- Always going up → monotone increasing
- Never drops below 2 → bounded below by 2
- Never goes above 3 → bounded above by 3
And because it’s increasing and doesn’t go to infinity (it’s capped at 3), it must settle toward some number — which is actually e ≈ 2.718.
That’s why the limit exists — and it’s called Euler’s number!
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Parent Tip: Review the logic above to help your child master the concept of calculus 2 worksheet.
