# How to find the limit of the sequence ((3n + 1)/(3n - 1))^n ?I've toyed around with this question for a bit but I can't come to any answer that seems right. I've seen somewhere that the answer...

How to find the limit of the sequence ((3n + 1)/(3n - 1))^n ?

I've toyed around with this question for a bit but I can't come to any answer that seems right. I've seen somewhere that the answer was e^(2/3). Not sure how that answer was arrived at.....

*print*Print*list*Cite

### 1 Answer

The typesetting may be a little unwieldy here, so please bear with me. One method is to do the long division to get

`(3n+1)/(3n-1)=1+2/(3n-1)` .

Then the nth term of our sequence is

`(1+2/(3n-1))^n` , which is starting to look similar to the definition of `e` in terms of the limit of the function `(1+1/x)^x` as `x->oo` , so we should try to keep going in this direction (also notice the 2/3 has popped up nicely). We would like the denominator and the exponent to be the same, so rewrite as

`(1+2/(3n-1))^n`

`=[1+2/3(1/(n-1/3))]^(n-1/3)*[1+2/3(1/(n-1/3))]^(1/3)`

As `n->oo` , clearly `[1+2/3(1/(n-1/3))]^(1/3)->1` , and if we make the substitution (to make the typesetting easier on the eyes) `h=n-1/3` , then `lim_(h->oo)(1+2/3(1/h))^h=e^(2/3)`

**Since the limit of a product is the product of the limits, we see that the limit is `e^(2/3)` **

**Sources:**