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I am studying for a Calc II exam and am confused by a fairly basic step with series and sequences. I understand the difference between the two but in all the book examples or online examples to discover if a series converges you are given a series.

On my professors study guide he gives us series and sequences and asks us to figure out if they converge/diverge.

How do you solve such a problem for a sequence. I will give you a problem from our study guide... This is not homework I really need this explained and want to try to figure out why for my exam.

The problem:

Determine whether the sequence converges or diverges and if it converges determine what it converges to.

$a_n = \frac{5n+6}{8n+3}$

We have tried a couple different tests but all the info for limit/ratio test are for series. Thanks for the help in advance


Added

So in the same light to determine if a series is convergent like

$\sum\limits_{x=2}^\infty \frac{2}{n^2-1}$

You see that it is harmonic... I think.

And you see that the bottom of the fraction grows to infinity you know that the series converges.

Then you can look at the limit of the entire thing $\frac{2}{n^2-1}$ and see that it converges to 0.

Is this thought process correct?

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Three things: (i) The series is not well-defined: when $n=1$, the denominator of the term is $0$; you want to start at $n=2$. And (ii) No, the series is not harmonic. The harmonic series is$$\sum_{n=1}^{\infty}\frac{1}{n},$$ and this is not that series. (iii) You are misapplying the Divergence Test. The Divergence Test says that if the terms do not converge to $0$, then the series diverges. But if the terms do converge to zero (as they do here), then this, by itself, does not tell you if the series converges or not. Your argument is invalid. –  Arturo Magidin Apr 20 '11 at 2:41
    
Whether a series converges or not is a question about what the sequence of partial sums does. Unfortunately, the sequence of partial sums is very hard to get a hold of in general; so instead, we try to deduce whether the series converges by looking at the sequence of terms. It's a bit like the drunk who is looking for his keys under the streetlamp, not because that's where he lost them, but because that's where there is enough light to search. That's why we have so many tests, and why many of them are often inconclusive. The Divergence Test can never tell you the series converges. –  Arturo Magidin Apr 20 '11 at 2:46
    
For extra credit, show that $${2\over n^2-1}={1\over n-1}-{1\over n+1}$$ and use this to compute $$\sum_{n=2}^{\infty}{2\over n^2-1}$$ –  Gerry Myerson Apr 20 '11 at 6:04

4 Answers 4

up vote 6 down vote accepted

As far as your added question goes, no, I'm afraid your thought process is invalid. You are essentially using a classic misapplication of the Divergence Test.

Divergence Test. If the limit of $a_n$ as $n\to\infty$ is not equal to $0$ (either does not exist, or exists and is not equal to $0$), then the series $$\sum a_n$$ diverges.

(Sometimes the Divergence Test is phrased in the contrapositive:

If $$\sum a_n$$ converges, then $\lim\limits_{n\to\infty}a_n = 0$.

but you are still trying to use it by affirming the consequent, which is a logical fallacy and an invalid mathematical argument). The Divergence Test can never let you conclude a series converges (look at the name: it's a dead giveaway). But that's what you are trying to do.

To go over the points I made in comments: whenever you have a series, $$\sum_{n=1}^{\infty}a_n$$ you automatically get two sequences that are associated to the series. The sequence that you are actually interested in, vis a vis the series, is the sequence of partial sums: $$\begin{align*} s_1 &= a_1\\ s_2 &= a_1+a_2\\ s_3 &= a_1+a_2+a_3\\ &\vdots\\ s_n &= a_1 + a_2 + \cdots + a_n\\ &\vdots \end{align*}$$ When we ask whether a series $\sum a_n$ converges, we are really asking whether the sequence $\{s_n\}$ of partial sums converges. When we say that the series "converges to $L$", $\sum a_n = L$, we are really saying that the sequence $s_n$ converges to $L$. When we say the series $\sum a_n$ diverges, we are really saying that the sequence of partial sums $\{s_n\}$ diverges.

Unfortunately, the sequence of partial sums is usually very hard to get a hold of. Thing about even a very simple series, like the harmonic series, $$\sum_{n=1}^{\infty}\frac{1}{n} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4}+\cdots + \frac{1}{n} + \cdots$$ The sequence of partial sums is pretty complicated, in terms of trying to get a formula for the $n$th term: $$\begin{align*} s_1 & = 1\\ s_2 &= 1 + \frac{1}{2} = \frac{3}{2}\\ s_3 &= 1 + \frac{1}{2}+\frac{1}{3} = \frac{11}{6}\\ s_4 &= 1 + \frac{1}{2}+\frac{1}{3}+\frac{1}{4} = \frac{25}{12}\\ &\vdots \end{align*}$$

The consequence of this difficulty is that we usually have a very hard time telling directly whether the sequence of partial sums converges, let alone what it converges to; because it's so hard to get a hold of the series.

So instead we look at the other sequence that we get from the series (which I alluded to above): when we have the series $$\sum_{n=1}^{\infty}a_n$$ we also get the sequence of terms, $$a_1,a_2,a_3,\ldots,a_n,\ldots$$ This is a much easier sequence to get a hold of, because we usually have it right in front of us.

Now, the sequence of terms is not what we are actually intereted in (we are actually interested in the sequence of partial sums), though it is of course related to the sequence of partial sums. Because the sequence of terms is so much easier to handle, though, we would like to be able to infer properties of the sequence of partial sums (in particular, whether it converges or not) from properties of the sequence of terms.

This is what pretty much all the tests for series that you see in Calculus II are about: they are ways in which you may be able to infer whether the sequence of partial sums converges or not, based on properties of the sequence of terms.

But, ultimately, we are looking in the wrong place: we are like the drunk searching for his keys under the streetlight, not because that's where the keys were lost, but because that's where there is enough light. Because we are really looking at the wrong place (and merely hoping to catch a glimpse of what we are really looking for from the corner of our eye), almost all tests are limited in some way: either you cannot always use them (e.g., the limit comparison test cannot be used if the series have positive and negative terms; the alternating series test cannot be used if the series is not alternating, etc); or they are not always conclusive (the Divergence Test can tell you a series diverges, but it can never tell you a series converges; the Ratio Test can be inconclusive; the Alternating Series Test can be inconclusive; etc). Because, ultimately, we are focusing on the wrong thing.

Which is what makes things so complicated and confusing for students. There is no recipe; instead, there are just a bunch of things you can try, and which may or may not work. You need to try several things, and you need to remember exactly what you can conclude and what you cannot from the different things you try, and when you can use them and when you cannot. It's a lot to keep in mind, but unfortunately it's the best we can do.

With practice, one comes to recognize certain series, to get more series "under one's belt" (so that we can do things like apply Comparison or Limit Comparison), or through experience get a feel for what kind of tests tend to work well with what kind of series (for instance, my experience tells me that trying to use the Ratio Test with the series you give in your addendum would be a waste of time, since it would come out inconclusive). But I'm afraid that comes with experience, and there is no clever mnemonic that you can memorize, or clever trick you can always use, or magic elixir you can drink, that will let you do it easily and every time (and without having to think about it too much). There just isn't any such thing, just like there isn't any such thing for finding antiderivatives.

For the series $$\sum_{n=2}^{\infty}\frac{1}{n^2-1},$$ what you did was apply the Divergence Test to see if the series might diverge; this is generally a good first step, because it is generally easy to do, and if the terms fail the divergence test (the terms do not go to $0$), then you are done: the series diverges. You performed the test correctly: it is indeed the case that $$\lim_{n\to\infty}\frac{1}{n^2-1} = 0$$ because the denominators grow without bound. Unfortunately, what this means is that the terms pass the Divergence Test, and therefore that the divergence test does not settle whether the series converges or diverges: the test is inconclusive.

Instead, you might note that this series is very close to the series $\sum\frac{1}{n^2}$; so you might want to try a comparison or limit comparison test with that series (assuming you know whether $\sum\frac{1}{n^2}$ converges or diverges).

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Really agree with this, having focused heavily on this topic in my studies this year. –  fdart17 Apr 20 '11 at 3:40

For this particular sequence, you could divide both numerator and denominator by $n$. Then $$a_n=\frac{5+\frac{6}{n}}{8+\frac{3}{n}}.$$ Taking the limit as $n\rightarrow \infty$ we can use the quotient property of limits to find that $$\lim_{n\rightarrow\infty }a_n=\frac{\lim_{n\rightarrow\infty } \left( 5+6\cdot\frac{1}{n}\right) }{\lim_{n\rightarrow\infty } \left(8+3\cdot\frac{1}{n} \right)}= \frac{5+6\cdot \lim_{n\rightarrow\infty } \frac{1}{n} }{8+3\cdot\lim_{n\rightarrow\infty } \frac{1}{n} }= \frac{5}{8}.$$ The last equality follows since $$\lim_{n\rightarrow \infty} \frac{1}{n}=0.$$ Alternatively we could prove that it converges from the definition, but if you are not already doing this in your course, it may be a bit complicated.

Hope that helps,

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How do you know to multiply by $\frac{n}{n}$ and then are you saying that it converges at $\frac{5}{8}$? –  austinbv Apr 20 '11 at 1:36
    
@zob: you're exploiting knowledge of the limiting behavior of the function $\frac1{x}$ at infinity. –  J. M. Apr 20 '11 at 1:40
    
@zobgib: Yes your sequence does converge to 5/8. The reason why I divided both numerator and denominator by $n$ is because then the main terms are constants, and also because dealing with $\frac{1}{n}$ as $n\rightarrow\infty$ is easy, as we know its limit. If we had say $$\frac{7n^2+2n+1}{5n^2+3n+2}$$ I would divide top and bottom by $n^2$, and we would find that the limit of this sequence is $\frac{7}{5}$. –  Eric Naslund Apr 20 '11 at 1:42
    
So, are you safe to say as a general rule if the main terms are constants you use a limit to find the convergence point and if the limit goes to infinity then the answer is divergent. –  austinbv Apr 20 '11 at 1:45

$${5n+6\over8n+3}={5n+6\over8n+3}\div{n\over n}={5+(6/n)\over8+(3/n)}$$ Now what happens as $n\to\infty$?

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A sequence is just a function whose domain is the natural numbers (that is, $1$, $2$, $3,\ldots$), or on some occasions the subset of all natural numbers greater than or equal to some number (e.g., "starting from $n=2$", etc).

So, first and foremost, sequences are functions. And like any function, you can ask about limits of this function. But because the domain of a sequence is such a special kind of set, there is really only one kind of limit for which the question "what is the limit of this sequence at $a$" makes sense: when we take the limit as $n\to\infty$ (because every other point of the domain is "isolated": there is no open interval containing the point on which the function is always defined).

So when we talk about the limit of a sequence, we are simply asking about its limit, as a function, as $n\to\infty$, and the meaning is the same as the meaning of any limit at $\infty$ for any function.

So, for example, when you are confronted with the question

What is the limit of the sequence $\displaystyle \frac{5n+6}{8n+3}$?

what you are being asked is to compute $$\lim_{n\to\infty}\frac{5n+6}{8n+3},$$ and all the usual things you know for functions apply (except things like L'Hopital's Rule, which requires functions to be differentiable, which sequences are not). So the same kind of limit "tricks' you know to handle, for example, $$\lim_{x\to\infty}\frac{5x+6}{8x+3}$$ also work for the sequence.

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You say L'Hopital's Rule rule does not apply but in this example L'Hopital's Rule gets you the correct answer, convergent at $\frac{5}{8}$ is that just coincidence? –  austinbv Apr 20 '11 at 1:46
    
@zobgib: In this particular case, because your sequence happens to be given by a continuous differentiable function, it works. But generally, sequences are not differentiable functions, because the limit that defines the derivative at a point makes no sense. If you had, for instance, $a_n = \frac{1}{n!}$, then it would be hard to express it as a continuous function to which you can apply L'Hopital's Rule (it can be done, but you have to invoke some complicated functions to do it). There are situations where you can use LHopital's (cont) –  Arturo Magidin Apr 20 '11 at 1:50
    
@zobgib: (cont) and situation where you cannot; unless you really understand limits and the hypothesis that go into being able to use L'Hopital's, it's probably best to avoid it when dealing with sequences, lest you try to use it in those occasions when you cannot. –  Arturo Magidin Apr 20 '11 at 1:51
    
@zob: in short, if you can help it, avoid l'Hôpital like the plague. –  J. M. Apr 20 '11 at 2:12
    
@J.M.: "... when doing limits of sequences." –  Arturo Magidin Apr 20 '11 at 2:13

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