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

Tara Alvarado

Tara Alvarado

Answered

2022-01-13

Verification of the limit of a succession
limn+ln(n+1)2n=0
Procedure.
|ln(n+1)2n|<ϵ

Answer & Explanation

braodagxj

braodagxj

Expert

2022-01-14Added 38 answers

Step 1
Let (an)n1 be a sequence defined by
an=ln(n+1)cn
cR+×
By definition,
ann+ellϵ>0NN,nN:
|anell|<ϵ
that is,
ln(n+1)cnn+0ϵ>0NN,nN:
|ln(n+1)cn0|<ϵ
Since that
|ln(n+1)cn0|=|ln(n+1)cn||n+1cn|=n+1cn
<ϵn>1c(ϵ1c)
Therefore,
Let ϵ>0 and NN such that n>1c(ϵ1c) with cR+×. Then for all nN such that nN, we have
|ln(n+1)cn0|=|ln(n+1)cn||n+1cn|=n+1cn
=1c+1cn<1c+ϵ1c=ϵ
Note: We know that for all xR+× we have that
aac{x1}{x}ln(x)x1
Reference: For all xR+× we have x1xln(x)x1

Annie Levasseur

Annie Levasseur

Expert

2022-01-15Added 30 answers

Step 1
First note
et1+t+12t2
t0
Let t=ln(1+n) and then n=et1
So
ln(1+n)n=tet1tt+12t2=11+12t
=22+ln(1+n)
For ϵ>0, letting
22+ln(1+n)<ϵ
gives
n>e2(1κ)κ1
Define
N=e2(1κ)κ+1
Then when nN, one has
|ln(1+n)n0|<ϵ
or
limnln(1+n)n=0

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get your answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?