Limit of Sequence: "2, $\frac{3}{2}, \frac{4}{3}, \frac{5}{4}$

In summary, The conversation discusses a limit problem involving finding the nth term of a sequence and the limit of the sequence. The use of l'hopital's rule is suggested, but there is confusion about how to determine the limit value from changing numbers. An online limit solver reveals the limit to be 1, achieved by taking the limit of the expression as n approaches infinity. The expert clarifies that the nth term should not be "cancelled out" but rather the limit is found by taking the limit of the expression as n approaches infinity.
  • #1
NoLimits
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Hello again,

I am having trouble with a particular limit problem and would appreciate any help/pointers you can offer. The question is asking for the nth term of the sequence [tex]2, \frac{3}{2}, \frac{4}{3}, \frac{5}{4}[/tex]

.. and also asks for a limit of the sequence. My immediate guess was to apply l'hopital's rule, which would mean setting n to approach infinity and using something like this:

[tex]lim_n→∞ \frac{n+1}{n}[/tex]

It seems to me like it could work, however I do not understand how an actual 'limit' value can be determined from a sequence of unknown and changing numbers ('n'). What I mean is, in order to make my limit work then the nth term would have to equal infinity, would it not?

** Edit **: According to an online limit solver the limit is 1, which I can see is possible if the n values are canceled out.
 
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  • #2
NoLimits said:
Hello again,

I am having trouble with a particular limit problem and would appreciate any help/pointers you can offer. The question is asking for the nth term of the sequence [tex]2, \frac{3}{2}, \frac{4}{3}, \frac{5}{4}[/tex]

.. and also asks for a limit of the sequence. My immediate guess was to apply l'hopital's rule, which would mean setting n to approach infinity and using something like this:

[tex]lim_n→∞ \frac{n+1}{n}[/tex]

It seems to me like it could work, however I do not understand how an actual 'limit' value can be determined from a sequence of unknown and changing numbers ('n'). What I mean is, in order to make my limit work then the nth term would have to equal infinity, would it not?

** Edit **: According to an online limit solver the limit is 1, which I can see is possible if the n values are canceled out.

You don't "cancel out" the ##n##'s. You write as$$
\frac{n+1} n = \frac n n + \frac 1 n = 1 + \frac 1 n$$and take the limit as ##n\to\infty## of that.
 
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1. What is the limit of this sequence?

The limit of this sequence is 1.

2. How can you determine the limit of a sequence?

The limit of a sequence can be determined by observing the pattern and behavior of the terms as n (the number of terms) approaches infinity. In this case, as n increases, the terms in the sequence approach 1.

3. Is this sequence convergent or divergent?

This sequence is convergent, as the terms approach a specific value (1) as n increases.

4. Can you provide a general formula for the nth term of this sequence?

Yes, the general formula for the nth term of this sequence is n/(n+1).

5. How is the limit of a sequence related to the concept of infinity?

The limit of a sequence is closely related to the concept of infinity, as it represents the value that the terms in the sequence approach as n (the number of terms) approaches infinity. In this case, as n increases, the terms in the sequence approach 1, which can be seen as a representation of infinity.

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