# Limit of Sequence a_n: Find Limit as n->∞

• merced
In summary, the conversation was about finding the limit of a sequence, specifically for the function a_n = 2^n/(3^n + 1). The speaker separated the fraction into (1/3)(2/3)^n and simplified it to limit = 1/3 lim_(n->infinity) (2/3)^n. They then asked for guidance on finding the limit of (2/3)^n, to which the other speaker explained that it is a general result that for any real number a between -1 and 1, the limit of a^n is zero. They also mentioned using the geometric series and the necessary criterion for a series to converge. The conversation ended with the speaker understanding and thanking the other speaker
merced
$$a_n = 2^n/(3^n + 1)$$

Find the limit as n -> infinity

So I separated the fraction into $$(1/3)(2/3)^n$$

Then limit = 1/3 $$lim_(n->infinity) (2/3)^n.$$

So, my question is, how do you find the limit $$(2/3)^n.$$ Is analysis of functions simply enough, i.e. looking at how fast they increase?

How did you transform a_n into $$(1/3)(2/3)^n$$?

Btw - is this for a real analysis class or some intro calculus class?

Oops, sorry, I guess I didn't look hard enough at my post.

It's really $$a_n = \frac{2^n}{3^{n+1}}$$

And it's Calculus II

cyanide-sun
Oh ok.

Didn't you see the general result that if -1<a<1, then $a^n\rightarrow 0$?

If not, and if you've seen a little bit of series theory, you can prove it this way: You know that the geometric series $\sum a^n$ converges for -1<a<1 and diverges otherwise. And you've also seen the basic necessary criterion for a series to converge, namely that the general term $a_n$ of any converging series goes to zero in the limit n-->infty.

So you can make use of these two fact by saying "I know that the geometric series of general term $a_n=a^n$ (where -1<a<1) converges, hence it must be that $a_n=a^n\rightarrow 0$ for -1<a<1."

Last edited:
Ok, I get what you're saying...but why is -1<a<1 here?

I proved that for any real number a btw -1 and 1, the limit of a^n is zero.

In particular, for a=2/3, we have that the limit of (2/3)^n is zero. Which is what you were wondering about.

Ooh, ok, I understand now.

Thanks!

## What is a limit of a sequence?

A limit of a sequence, denoted as limn→∞ an, is the value that the terms of the sequence approach as n (the index of the terms) approaches infinity. In other words, it is the value that the terms of the sequence get closer and closer to as n increases without bound.

## How do you find the limit of a sequence?

To find the limit of a sequence, you can use two methods: the algebraic method and the graphical method. The algebraic method involves finding a formula for the n-th term of the sequence, then taking the limit as n approaches infinity. The graphical method involves plotting the terms of the sequence on a graph and observing the trend as n increases.

## Why is finding the limit of a sequence important?

Finding the limit of a sequence is important because it helps us understand the behavior of the sequence and determine if it converges (approaches a finite value) or diverges (does not approach a finite value). It also allows us to make predictions about the long-term behavior of the sequence.

## What is the difference between a finite and infinite limit of a sequence?

A finite limit of a sequence means that the terms of the sequence approach a specific finite value as n approaches infinity. This indicates that the sequence converges. On the other hand, an infinite limit of a sequence means that the terms of the sequence do not approach a specific finite value and continue to increase or decrease without bound. This indicates that the sequence diverges.

## What are some common misconceptions about limits of sequences?

One common misconception is that a sequence must approach a specific value in order to have a limit. In reality, a sequence can also approach infinity or negative infinity and still have a limit. Another misconception is that a sequence must have a limit in order to be a valid sequence, when in fact some sequences may not have a limit at all.

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