# Sequences convergence or not

• June 24th 2012, 02:26 PM
LAPOSH42
Sequences convergence or not
For the sequence below, determine whether they converge or not, and find the limit when the sequence converges

a)
an=4^n/n!
a=n+1/an=4/n+1 a=n+1 <a=n convergent

b)an=nsin(1/n)
an=sin(1/n)/(1/n) =1 convergent

an=(3/2)n
3/2 >1 divergent
• June 24th 2012, 06:16 PM
HallsofIvy
Re: Sequences convergence or not
Quote:

Originally Posted by LAPOSH42
For the sequence below, determine whether they converge or not, and find the limit when the sequence converges

a)
an=4^n/n!
a=n+1/an=4/n+1 a=n+1 <a=n convergent

This is really very confusingly written. I think you mean an+1 where you wrote "a= n+1". That is, you are using the ratio test: $\frac{a_{n+1}}{a_n}= \frac{4^{n+1}}{(n+1)!}\frac{n!}{4^n}= \frac{4}{n+1}$ which goes to 0< 1 as n goes to infinity so the series converges.
(But "an+1< an alone does NOT imply $\sum a_n$ converges.)

As for what it converges to, [itex]\sum_{n=0}^\infty \frac{x^n}{n!}= e^x[/tex].

Quote:

b)an=nsin(1/n)
an=sin(1/n)/(1/n) =1 convergent
What??? sin(1/n)/(1/n) is certainly NOT equal to 1! And even if it were, that would NOT prove it was convergent. What are you trying to say here?

Quote:

an=(3/2)n
3/2 >1 divergent
Yes, this is a geometric series $\sum r^n$ with r> 1 and so diverges.
• June 24th 2012, 06:56 PM
Reckoner
Re: Sequences convergence or not
Ivy, I think the original poster has to determine whether the sequences converge, not the corresponding series. I think he (or she) has got the right idea for the first two problems - showing that the sum of a sequence converges is sufficient to show that the sequence converges (to zero), and the second problem involves evaluating $\textstyle\lim_{n\to\infty}\frac{\sin 1/n}{1/n}.$ I do agree, however, that this is not at all clear with the confusing notation that the poster is using.