Is showing limz→∞(1+1z)z exists the same as limn→∞(1+1n)n exists

Let's denote an=(1+1n)n for the members of the sequence. Because 0&lt;n≤x&lt;n+1, then 1&lt;(1+1n+1)&lt;(1+1x)≤(1+1n). Consequently we get the upper bound (1+1x)x≤(1+1x)n+1&lt;(1+1n)n+1=an⋅(1+1n), and similarly the lower bound (1+1x)x≥(1+1n+1)n=an+1⋅(1+1n+1)−1. So (1+1x)x is sandwiched between two consecutive entries of the sequence (an) multiplied by terms →1 To answer: Yes, the two limit problems are equivalent.

By the definition of limits. c=limz→∞(1+1z)z exists if and only if ∀(sn)n∈N,(limn→∞(sn)=+∞)⇒ (s=limn→∞(1+1sn)sn exists and s=c).

Let 0&lt;x1&lt;x2. Then 0&lt;x1x2&lt;1, so by the Bernoullis

Let's denote an=(1+1n)n for the members of the sequence. Because 0&lt;n≤x&lt;n+1, then 1&lt;(1+1n+1)&lt;(1+1x)≤(1+1n). Consequently we get the upper bound (1+1x)x≤(1+1x)n+1&lt;(1+1n)n+1=an⋅(1+1n), and similarly the lower bound (1+1x)x≥(1+1n+1)n=an+1⋅(1+1n+1)−1. So (1+1x)x is sandwiched between two consecutive entries of the sequence (an) multiplied by terms →1 To answer: Yes, the two limit problems are equivalent.

By the definition of limits. c=limz→∞(1+1z)z exists if and only if ∀(sn)n∈N,(limn→∞(sn)=+∞)⇒ (s=limn→∞(1+1sn)sn exists and s=c).

Let 0&lt;x1&lt;x2. Then 0&lt;x1x2&lt;1, so by the Bernoullis

Get the answer to "Is showing limz→∞(1+1z)z exists the same as limn→∞(1+1n)n..."

pogonofor9z

Answered question

2022-01-20

Is showing

lim_{z \to \infty} {(1 + \frac{1}{z})}^{z}

exists the same as

lim_{n \to \infty} (1 + \frac{1}{n})

n exists

Answer & Explanation

hysgubwyri3

Beginner2022-01-20Added 43 answers

Let's denote $a_{n} = {(1 + \frac{1}{n})}^{n}$ for the members of the sequence.
Because $0 < n \leq x < n + 1,$ then
$1 < (1 + \frac{1}{n + 1}) < (1 + \frac{1}{x}) \leq (1 + \frac{1}{n}) .$
Consequently we get the upper bound
${(1 + \frac{1}{x})}^{x} \leq {(1 + \frac{1}{x})}^{n + 1} < {(1 + \frac{1}{n})}^{n + 1} = a_{n} \cdot (1 + \frac{1}{n}),$
and similarly the lower bound
${(1 + \frac{1}{x})}^{x} \geq {(1 + \frac{1}{n + 1})}^{n} = a_{n + 1} \cdot {(1 + \frac{1}{n + 1})}^{- 1} .$
So $(1 + \frac{1}{x})^{x}$ is sandwiched between two consecutive entries of the sequence $(a_{n})$ multiplied by terms $\to 1$
To answer: Yes, the two limit problems are equivalent.

Joseph Fair

Beginner2022-01-21Added 34 answers

By the definition of limits.

c = lim_{z \to \infty} {(1 + \frac{1}{z})}^{z}

exists if and only if

\forall (s_{n}) n \in N, (lim n \to \infty (s_{n}) = + \infty) \Rightarrow

(

s = lim_{n \to \infty} {(1 + \frac{1}{s_{n}})}^{s_{n}}

exists and

s = c

RizerMix

Expert2022-01-27Added 656 answers

Let

0 < x_{1} < x_{2} .

Then

0 < \frac{x_{1}}{x_{2}} < 1

, so by the Bernoullis

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