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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

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    You’re confusing sequences and series in your terminology. Are you talking about the **sequence** $\frac32,\frac43,\frac54,\dots$ whose $n$-th term is $\frac{n+2}{n+1}$, or are you talking about a **series**, an infinite sum of terms?2012-11-03
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    Yes, that maybe because I am a little confused. I am talking about the sequence. And how to find the limit.. Thanks2012-11-03

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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 $1$.

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    Thank you for your answer, I am still confused as to how how we find the limit, but I think I may be able to work it out from your answer2012-11-03
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    Also, when you say the limit is 1, do you mean that the sequence will never rise above zero.. thanks.. I am teaching myself mathematics, so I find some things confusing2012-11-03
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    When one says that the limit is $1$, one is saying, informally, that when $n$ is very big then $\frac{n+2}{n+1}$ is awfully close to $1$.2012-11-03
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    thank you, so much. I actually understand what going on a little better now2012-11-03
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For all when $x$ approches infinity 1) If degree is equal on up and down then answer will be equal to the divison of both(up nd down) with constants. Example. $\lim_{x\to\inf\frac{n+2}{n+1}}$ both up nd down are power with $1$ then $1/1=1$

2). If degree of variable is greater on up than degree of down then answer will be $+$ or minus infinity.

3). If degree of variable is less on up than degree of down then answer will be zero.

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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

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    It's difficult to grasp what you mean about sequences and "values of n falling in this lane". This is an older Question, so don't rush your Answer. See the [Help Center](http://math.stackexchange.com/help) for how to write good Answers, and be aware that this site allows the use of LaTeX formatting via [MathJax](http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference).2014-09-08