Let an=(1+(1)/(n))^n. (a) Show that if 0 ⩽a<b, then (b^n+1-a^n+1)/(b-a)<(n+1) b^n (b) Deduce tha

Let an=(1+(1)/(n))^n. (a) Show that if 0 ⩽a<b, then...

Let $a_n=\left(1+\frac{1}{n}\right)^n$. (a) Show that if $0 \leqslant a<b$, then $$ \frac{b^{n+1}-a^{n+1}}{b-a}<(n+1) b^n $$ (b) Deduce that $b^n[(n+1) a-n b]<a^{n+1}$. (c) Use $a=1+1 /(n+1)$ and $b=1+1 / n$ in part (b) to show that $\left\{a_n\right\}$ is increasing. (d) Use $a=1$ and $b=1+1 /(2 n)$ in part (b) to show that $a_{20}<4$. (e) Use parts (c) and (d) to show that $a_n<4$ for all $n$. (f) Use Theorem 12 to show that $\lim _{n \rightarrow \infty}(1+1 / n)^n$ exists. (The limit is $e$. See Equation 6.4.9 or 6.4*.9.)

Let $a_n=\left(1+\frac{1}{n}\right)^n$.
(a) Show that if $0 \leqslant a<b$, then

$$
\frac{b^{n+1}-a^{n+1}}{b-a}<(n+1) b^n
$$

(b) Deduce that $b^n[(n+1) a-n b]<a^{n+1}$.
(c) Use $a=1+1 /(n+1)$ and $b=1+1 / n$ in part (b) to show that $\left\{a_n\right\}$ is increasing.
(d) Use $a=1$ and $b=1+1 /(2 n)$ in part (b) to show that $a_{20}<4$.
(e) Use parts (c) and (d) to show that $a_n<4$ for all $n$.
(f) Use Theorem 12 to show that $\lim _{n \rightarrow \infty}(1+1 / n)^n$ exists. (The limit is $e$. See Equation 6.4.9 or 6.4*.9.)

Key Concepts

Bounded Sequences

A bounded sequence is one for which there exists a real number that serves as an upper (or lower) limit that no term of the sequence exceeds. Boundedness plays a crucial role in proving convergence, because a sequence that is both bounded and monotonic is guaranteed to converge by the Monotone Convergence Theorem.

Sequences and Limits

A sequence is an ordered list of numbers, and the limit of a sequence is the value that its terms approach as the index increases indefinitely. Understanding limits is essential in analysis, as it helps determine the behavior of sequences and underlies many concepts, such as continuity, the definition of the exponential function, and convergence properties.

Monotonic Sequences

A monotonic sequence is one that is either entirely non-decreasing or non-increasing. Establishing that a sequence is monotonic can simplify the analysis of its convergence properties since monotonicity often allows us to use powerful theorems to confirm convergence, particularly when combined with boundedness.

Monotone Convergence Theorem

The Monotone Convergence Theorem states that every sequence that is monotonic and bounded converges. This theorem is a fundamental tool in real analysis, allowing one to establish the existence of limits for sequences by confirming that they are both monotonic (either increasing or decreasing) and bounded.

Exponential Functions and the Limit Definition of e

The exponential function is key in many areas of mathematics, and one of its fundamental representations is given by the limit of the sequence (1 + 1/n)^n as n tends to infinity. This limit is defined to be the number e, which serves as the base of the natural logarithm and has deep implications in calculus, differential equations, and complex analysis.

Transcript

00:01 So for the first part, a, we have b to the power n plus 1 minus a to the power n plus 1 divided by b minus a.

00:17 And this is equal to b to the power n plus b to the power n minus 1 a plus b to the power n minus 1 a plus b to the power n minus 2.

00:32 Plus b to the power n minus 3 a key plus 2 the nth serve that is b a n minus 1 plus a to the power n and this is less than b to the power n plus b to the power n plus b to the power n plus b to the power n plus b to the power power n minus 2 b squared plus b to the power n minus 3 b plus plus b to the power n minus 1 plus b to the power n and this is equal to n plus n plus the power n.

01:40 Then for b since b minus a is greater than zero it implies that we have b to the power n minus 1 minus a to be less than n plus 1 to the power b to the power n times b minus a so this implies that b to the power n minus one minus n plus one n plus one b to the power n we have b minus a is less than a to the power n plus one so this implies that we have b to the power n plus one so this implies that we have b to the power n plus one.

02:45 We have n plus 1 a minus n p to be less than a to the power n plus 1 then for c the third part with substitution we have n plus 1 we have n plus 1 a plus 1 b to be equal to 1 so then it implies that b to the power n it's equal to 1 plus 1 divided by n to the power n less than a to the power n plus 1 and this is equal to 1 plus 1 and this is equal to 1 plus 1 n n plus 1 then for d the fourth part which is d with substitution substitution we have 1 plus 1 divided by 2 n to the power n times 1 divided by 2n to be less than 1 so this implies that one plus one divided by two n to the power n will be less than two so then we have we have one plus one divided by two n to the power two n to be less than four when you square both sides then for e, for a .n less than a to n.

05:26 So a .n is less than this.

05:29 Since our sequence, a .n is increasing.

05:42 So this, therefore, a .n is less than a to n less than four...

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