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
Best Answer
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 $1$.