Verification of the limit of a succession limn→+∞ln⁡(n+1)2n=0 Procedure. |ln⁡(n+1)2n|&lt;ϵ

Step 1 Let (an)n≥1 be a sequence defined by an=ln⁡(n+1)cn c∈R+× By definition, an∼n→+∞ell⇔∀ϵ&gt;0∃N∈N,n≥N: |an−ell|&lt;ϵ that is, ln⁡(n+1)cn∼n→+∞0⇔∀ϵ&gt;0∃N∈N,n≥N: |ln⁡(n+1)cn−0|&lt;ϵ Since that |ln⁡(n+1)cn−0|=|ln⁡(n+1)cn|≤|n+1cn|=n+1cn &lt;ϵ⇔n&gt;1c(ϵ−1c) Therefore, Let ϵ&gt;0 and N∈N such that n&gt;1c(ϵ−1c) with c∈R+×. Then for all n∈N such that n≥N, we have |ln⁡(n+1)cn−0|=|ln⁡(n+1)cn|≤|n+1cn|=n+1cn =1c+1cn&lt;1c+ϵ−1c=ϵ Note: We know that for all x∈R+× we have that aac{x−1}{x}≤ln⁡(x)≤x−1 Reference: For all x∈R+× we have x−1x≤ln⁡(x)≤x−1

Step 1 First note et≥1+t+12t2 t≥0 Let t=ln⁡(1+n) and then n=et−1 So ln⁡(1+n)n=tet−1≤tt+12t2=11+12t =22+ln⁡(1+n) For ∀ϵ&gt;0, letting 22+ln⁡(1+n)&lt;ϵ gives n&gt;e2(1−κ)κ−1 Define N=⌊e2(1−κ)κ⌋+1 Then when n≥N, one has |ln⁡(1+n)n−0|&lt;ϵ or limn→∞ln⁡(1+n)n=0

Step 1 Let (an)n≥1 be a sequence defined by an=ln⁡(n+1)cn c∈R+× By definition, an∼n→+∞ell⇔∀ϵ&gt;0∃N∈N,n≥N: |an−ell|&lt;ϵ that is, ln⁡(n+1)cn∼n→+∞0⇔∀ϵ&gt;0∃N∈N,n≥N: |ln⁡(n+1)cn−0|&lt;ϵ Since that |ln⁡(n+1)cn−0|=|ln⁡(n+1)cn|≤|n+1cn|=n+1cn &lt;ϵ⇔n&gt;1c(ϵ−1c) Therefore, Let ϵ&gt;0 and N∈N such that n&gt;1c(ϵ−1c) with c∈R+×. Then for all n∈N such that n≥N, we have |ln⁡(n+1)cn−0|=|ln⁡(n+1)cn|≤|n+1cn|=n+1cn =1c+1cn&lt;1c+ϵ−1c=ϵ Note: We know that for all x∈R+× we have that aac{x−1}{x}≤ln⁡(x)≤x−1 Reference: For all x∈R+× we have x−1x≤ln⁡(x)≤x−1

Step 1 First note et≥1+t+12t2 t≥0 Let t=ln⁡(1+n) and then n=et−1 So ln⁡(1+n)n=tet−1≤tt+12t2=11+12t =22+ln⁡(1+n) For ∀ϵ&gt;0, letting 22+ln⁡(1+n)&lt;ϵ gives n&gt;e2(1−κ)κ−1 Define N=⌊e2(1−κ)κ⌋+1 Then when n≥N, one has |ln⁡(1+n)n−0|&lt;ϵ or limn→∞ln⁡(1+n)n=0

Expert Answer to "Verification of the limit of a succession limn→+∞ln⁡(n+1)2n=0..."

Tara Alvarado

Answered question

2022-01-13

Verification of the limit of a succession

lim_{n \to + \infty} \frac{\ln (n + 1)}{2 n} = 0

Procedure.

| \frac{\ln (n + 1)}{2 n} | < ϵ

Answer & Explanation

braodagxj

Beginner2022-01-14Added 38 answers

Step 1
Let ${(a_{n})}_{n \geq 1}$ be a sequence defined by
$a_{n} = \frac{\ln (n + 1)}{c n}$
$c \in R_{+}^{\times}$
By definition,
$a_{n} \sim_{n \to + \infty} e l l \Leftrightarrow \forall_{ϵ} > 0 \exists N \in N, n \geq N :$
$| a_{n} - e l l | < ϵ$
that is,
$\frac{\ln (n + 1)}{c n} \sim_{n \to + \infty} 0 \Leftrightarrow \forall_{ϵ} > 0 \exists N \in N, n \geq N :$
$| \frac{\ln (n + 1)}{c n} - 0 | < ϵ$
Since that
$| \frac{\ln (n + 1)}{c n} - 0 | = | \frac{\ln (n + 1)}{c n} | \leq | \frac{n + 1}{c n} | = \frac{n + 1}{c n}$
$< ϵ \Leftrightarrow n > \frac{1}{c (ϵ - \frac{1}{c})}$
Therefore,
Let $ϵ > 0$ and $N \in N$ such that $n > \frac{1}{c (ϵ - \frac{1}{c})}$ with $c \in R_{+}^{\times}$ . Then for all $n \in N$ such that $n \geq N$ , we have
$| \frac{\ln (n + 1)}{c n} - 0 | = | \frac{\ln (n + 1)}{c n} | \leq | \frac{n + 1}{c n} | = \frac{n + 1}{c n}$
$= \frac{1}{c} + \frac{1}{c n} < \frac{1}{c} + ϵ - \frac{1}{c} = ϵ$
Note: We know that for all $x \in R_{+}^{\times}$ we have that
$a a c {x - 1} {x} \leq \ln (x) \leq x - 1$
Reference: For all $x \in R_{+}^{\times}$ we have $\frac{x - 1}{x} \leq \ln (x) \leq x - 1$

Annie Levasseur

Beginner2022-01-15Added 30 answers

Step 1
First note
$e^{t} \geq 1 + t + \frac{1}{2} t^{2}$
$t \geq 0$
Let $t = \ln (1 + n)$ and then $n = e^{t} - 1$
So
$\frac{\ln (1 + n)}{n} = \frac{t}{e^{t} - 1} \leq \frac{t}{t + \frac{1}{2} t^{2}} = \frac{1}{1 + \frac{1}{2} t}$
$= \frac{2}{2 + \ln (1 + n)}$
For $\forall_{ϵ} > 0$ , letting
$\frac{2}{2 + \ln (1 + n)} < ϵ$
gives
$n > e^{\frac{2 (1 - κ)}{κ}} - 1$
Define
$N = ⌊ e^{\frac{2 (1 - κ)}{κ}} ⌋ + 1$
Then when $n \geq N$ , one has
$| \frac{\ln (1 + n)}{n} - 0 | < ϵ$
or
$lim_{n \to \infty} \frac{\ln (1 + n)}{n} = 0$

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