Sequences convergence or not

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

a)

a_{n}=4^n/n!

a=n+1/a_{n}=4/n+1 a=n+1 <a=n convergent

b)a_{n}=nsin(1/n)

a_{n}=sin(1/n)/(1/n) =1 convergent

a_{n}=(3/2)^{n}

3/2 >1 divergent

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)

a_{n}=4^n/n!

a=n+1/a_{n}=4/n+1 a=n+1 <a=n convergent

This is really very confusingly written. I **think** you mean a_{n+1} where you wrote "a= n+1". That is, you are using the ratio test: $\displaystyle \frac{a_{n+1}}{a_n}= \frac{4^{n+1}}{(n+1)!}\frac{n!}{4^n}= \frac{4}{n+1}$ which goes to 0< 1 as n goes to infinity so the series converges.

(But "a_{n+1}< a_{n} alone does NOT imply $\displaystyle \sum a_n$ converges.)

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

Quote:

b)a_{n}=nsin(1/n)

a_{n}=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:

a_{n}=(3/2)^{n}

3/2 >1 divergent

Yes, this is a geometric series $\displaystyle \sum r^n$ with r> 1 and so diverges.

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 $\displaystyle \textstyle\lim_{n\to\infty}\frac{\sin 1/n}{1/n}.$ I do agree, however, that this is not at all clear with the confusing notation that the poster is using.