Sequence And Series - How To Find The Limit Of A Converging Sequence With No Common Ratio I have the following series 
3/2, 4/3, 5/4, 6/5....

the nth term could be expressed as n+2/n+1.  

I cant seem to work out how to get the limit of this function as most of the books describe the sum to infinity as a/1 -r.

With a = 3/2 and being the first term, I am confused as to what r should equal.  As this sequence does not seem to have a common ratio.  I thought that r would be the common ratio.  

Maybe I am going down the wrong line of thinking so please clarify how to find the limit and the r variable in the equation.

Many Thanks
Suppose that the $n$-th term of a certain sequence is $\dfrac{n+2}{n+1}$.  Note that 
$$\frac{n+2}{n+1}=1+\frac{1}{n+1}.$$
 As $n\to\infty$, the $\dfrac{1}{n+1}$ part approaches $0$, so our limit is $1$.

Or else we can divide top and bottom by $n$, obtaining
$$\frac{n+2}{n+1}=\frac{1+\frac{2}{n}}{1+\frac{1}{n}}.$$
As $n\to\infty$, $1+\dfrac{2}{n}\to 1$, and $1+\dfrac{1}{n}\to 1$, so our limit is $1$.

Remark: One can get useful information from the calculator. If you are interested in finding the limit as $n\to\infty$ of $a_n$, you can calculate $a_n$ for a few largish $n$. For example, ca;culate $\dfrac{n+2}{n+1}$ for $n=1000$ and for $n=10000$. It will now be plausible that the limit might be $1$. And once one knows that the "answer" should be $1$, it becomes easier to show that it is $1$.
for a seaquency Un..a limit L is such that for its value there is a positive integer N such that n>N..so a sequency only has this limit L if there are still some integer values of n falling in this lane