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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? –  Brian M. Scott Nov 3 '12 at 18:42
    
Yes, that maybe because I am a little confused. I am talking about the sequence. And how to find the limit.. Thanks –  user866190 Nov 3 '12 at 18:47

3 Answers 3

up vote 3 down vote accepted

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 answer –  user866190 Nov 3 '12 at 18:50
    
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 confusing –  user866190 Nov 3 '12 at 18:53
    
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 $1$. –  André Nicolas Nov 3 '12 at 18:56
    
thank you, so much. I actually understand what going on a little better now –  user866190 Nov 3 '12 at 18:58

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 for how to write good Answers, and be aware that this site allows the use of LaTeX formatting via MathJax. –  hardmath Sep 8 at 18:48

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