Given:Sigma, n=1 going up to infinity.How would you test whether this is convergence or divergence?(n+3)/(n+2)My text book doesn't explain it to me properly and leaves me more confused.

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- Oct 1st 2012, 04:41 AMMathIsOhSoHardHow do you test for convergence or divergence?
Given:Sigma, n=1 going up to infinity.How would you test whether this is convergence or divergence?(n+3)/(n+2)My text book doesn't explain it to me properly and leaves me more confused.

- Oct 1st 2012, 06:09 AMPlatoRe: How do you test for convergence or divergence?
The series $\displaystyle \sum\limits_{n = 0}^\infty {a_n } $ converges only if $\displaystyle \left( {a_n } \right) \to 0$.

That means that if $\displaystyle \left( {a_n } \right)\not \to 0$ then the series diverges.

What about $\displaystyle \left( {\frac{n+3}{n+2} } \right)\to ?$ - Oct 1st 2012, 06:20 AMMathIsOhSoHardRe: How do you test for convergence or divergence?
My guess is that the fraction can be reduced to 3/2 which means that a_n is equal to 3/2.

Since it does not converge towards 0, the fraction is divergence.

But would it be correct to say the fraction is equal to 3/2 or would it be correct to say the fraction goes towards 3/2? Im kinda thinking that since the fraction does not contain any n value that goes towards infinity, then the correct term would be "equal" and not "towards"? - Oct 1st 2012, 06:36 AMPlatoRe: How do you test for convergence or divergence?
You are really confused on terminology.

Because the**sequence**$\displaystyle \left( {\frac{{{\rm{n + 3}}}}{{{\rm{n + 2}}}}} \right) \to 1$ the**series**$\displaystyle \sum\limits_{n = 1}^\infty {\left( {\frac{{{\rm{n + 3}}}}{{{\rm{n + 2}}}}} \right)} $ diverges. - Oct 1st 2012, 06:45 AMDevenoRe: How do you test for convergence or divergence?
here is what we have:

$\displaystyle \frac{n+3}{n+2} = \frac{(n+2) + 1}{n+2} = 1 + \frac{1}{n+2} > 1$.

therefore:

$\displaystyle \sum_{n=1}^k a_n = \sum_{n=1}^k \frac{n+3}{n+2} > k$.

thus given ANY integer N, we can ALWAYS find an integer k such that: $\displaystyle \sum_{n=1}^k \frac{n+3}{n+2}> N$ (k = N will work nicely).

thus the**series**$\displaystyle \sum_{n=1}^\infty \frac{n+3}{n+2} $ diverges because the**sequence**of partial sums {S_{k}}:

$\displaystyle S_k = \sum_{n=1}^k \frac{n+3}{n+2} $ increases without bound. - Oct 1st 2012, 09:03 AMMathIsOhSoHardRe: How do you test for convergence or divergence?
I'm a bit confused now because to me it seems like Plato is saying the sequence goes towards 1 but Deveno is saying it increases without bound.

I think there must be something I'm missing.

But we have the definition of the partial sums as:

$\displaystyle S_N=\frac{1+3}{1+2}+\frac{2+3}{2+2}+...+\frac{N+3} {N+2}=\sum_{n=1}^N \frac{n+3}{n+2}$

So the partial sums is the sum of the sequence up to $\displaystyle N$. Is that correct?

If we can determine that the sequences are convergent, then the series is convergent as well or vice versa (if the sequence is divergent then the series is divergent.)

The nth term test tells us that in**series**if $\displaystyle a_n$ does not go towards 0, then the series is divergent (so if $\displaystyle a_n$ goes towards 0 then the series is convergent).

Am I on the right track? So I need to figure out whether the sequence is convergent or not, and then I can use that information to conclude if the series is convergent or not. - Oct 1st 2012, 09:21 AMPlatoRe: How do you test for convergence or divergence?
That is totally and completely false.

You are very confused on the vocabulary.

Series are about the convergence of partials sums.

Partials sums are finite sums of some sequence.

Now why is that false?

The sequence $\displaystyle \left( {\frac{1}{n}} \right) \to 0$ BUT $\displaystyle \sum\limits_{n = 1}^\infty {\left( {\frac{1}{n}} \right)} $ DIVERGES! - Oct 1st 2012, 10:06 AMMathIsOhSoHardRe: How do you test for convergence or divergence?
If I remember correctly, $\displaystyle \frac{1}{n}$ is a harmonic function (not sure if that is the correct term but something "harmonic").

So the series for this harmonic function, starting with n=1, becomes:

$\displaystyle 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}...$

We see in this series that it already starts at 1, and then it keeps getting added to it, so it surpasses 1.

Going by the nth term test, which says that if $\displaystyle a_n$ does not go towards 0, then the series is divergent.

This is the case here, since it is way over 1 and for every sequence more gets added to it making it go further away. So we can conclude by the nth term test that it is in fact divergent.

This is only for the**series**though!

Because the series is a sum where you add all the sequences together.

But for the**sequence**, we don't add all the previous together. Here $\displaystyle n\rightarrow\infty$ so the fraction becomes smaller and smaller as $\displaystyle n$ becomes larger and larger.

The difference between the sequence and the series that you mentioned is that the sequence is only one fraction where n becomes infinitely small and the series is a set of inifinite fractions that are all added together where n each time becomes 1 integer bigger.

I really hope some of it was correct lol

And you're right, the nth term test does not show us whether a series is convergent or not.

It can only be used to show us if a series is divergent (we need other types of tests to show the series is convergent). - Oct 2nd 2012, 12:18 PMDevenoRe: How do you test for convergence or divergence?
yes. in fact, convergence of a series can be VERY hard to show. the reason is a convergent (or divergent) series can converge (or diverge) very slowly.

however we do NOT "add all the sequences together".

there are 3 separate things involved:

1. a sequence of TERMS $\displaystyle \{a_n\}$.

2. a sequence of "partial sums" (adding all the terms up to the k-th term} $\displaystyle \{S_k\},\ S_k = \sum_{n=1}^k a_n$.

3. the series itself: $\displaystyle \sum_{n=1}^\infty a_n = \lim_{k \to \infty} S_k$.

now, if the limit in number 3 exists, the partial sums need to get "closer and closer together": $\displaystyle \lim_{k \to \infty} S_k - S_{k-1} = 0$.

but what is $\displaystyle S_k - S_{k-1}$? it's just the k-th term, $\displaystyle a_k$.

so: IF a series is convergent (if the sequence of partial sums converges), the limit of the terms is 0.

however, just because the terms "go to 0" doesn't mean the series converges. the prime example of this is (as you said) the harmonic series:

$\displaystyle \sum_{n=1}^\infty \frac{1}{n}$.

it is clear that $\displaystyle \lim_{n \to \infty} \frac{1}{n} = 0$.

however, let's look at the "$\displaystyle 2^k$-th" partial sum:

$\displaystyle \sum_{n = 1}^{2^k} \frac{1}{n}$

$\displaystyle = 1 + \frac{1}{2} + \left(\frac{1}{3} + \frac{1}{4}\right) + \left(\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8}\right) + \cdots + \left(\frac{1}{2^{k-1}} + \cdots + \frac{1}{2^k}\right)$

$\displaystyle > \frac{1}{2} + \frac{1}{2} + \left(\frac{1}{4} + \frac{1}{4}\right) + \left(\frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8}\right) + \cdots + \left(\frac{1}{2^k} + \cdots + \frac{1}{2^k}\right) = \frac{k+1}{2}$.

so, if we pick any positive integer N, and we pick k such that k > 4^{N}, then:

$\displaystyle S_k > S_{4^N} = S_{2^{(2N)}} > \frac{2N + 1}{2} > \frac{2N}{2} = N$

which shows that $\displaystyle \lim_{k \to \infty} S_k$ does not exist. - Oct 2nd 2012, 03:51 PMMathIsOhSoHardRe: How do you test for convergence or divergence?
I see, that does make sense, thanks!

But one thing that I'm still not sure about is the following example:

$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n(n+1)}$

This series is convergent and the sequence of partial sums is $\displaystyle S_k\to 1$

But how come this following series:

$\displaystyle \sum_{n=1}^{\infty}\frac{1}{2(n+2)}$

diverges?

What is the difference between the two of them and what convergence test would you use to prove the first series convergent and the second series divergent? - Oct 2nd 2012, 09:31 PMDevenoRe: How do you test for convergence or divergence?
one way to tell if a series is convergent is: if it is "dominated" by another series, that is:

$\displaystyle 0 \leq a_n \leq b_n$ for all n, and $\displaystyle \sum_{n=1}^\infty b_n$ converges. note that this only works if every term is non-negative.

by the same token, if some series dominates a divergent series, it must also be divergent.

now $\displaystyle \frac{1}{n(n+1)} < \frac{1}{n^2}$ so if:

$\displaystyle \sum_{n=1}^\infty \frac{1}{n^2}$ converges, so does your first series. it is a well-known result that the latter series does indeed converge to $\displaystyle \frac{\pi^2}{6}$.

but that doesn't tell us what the first series converges TO. however, since i "know the answer" i can prove what it converges to.

namely, i will prove by induction on m that:

$\displaystyle \sum_{n=1}^m \frac{1}{n(n+1)} = \frac{m}{m+1}$. this is clearly true for m = 1. assume it's true for m = k. then:

$\displaystyle \sum_{n=1}^{k+1} \frac{1}{n(n+1)} = \left(\sum_{n=1}^k \frac{1}{n(n+1)}\right) + \frac{1}{(k+1)(k+2)}$

$\displaystyle = \frac{k}{k+1} + \frac{1}{(k+1)(k+2)} = \frac{k(k+2)}{(k+1)(k+2)} + \frac{1}{(k+1)(k+2)}$

$\displaystyle = \frac{k^2 + 2k + 1}{(k+1)(k+2)} = \frac{k+1}{k+2}$, which is the assertion for m = k+1.

hence $\displaystyle \lim_{m \to \infty} S_m = \lim_{m \to \infty} \frac{m}{m+1} = \lim_{m \to \infty} 1 - \lim_{m \to \infty}\frac{1}{m+1} = 1 - 0 = 1$, so:

$\displaystyle \sum_{n=1}^\infty \frac{1}{n(n+1)} = 1$.

***************************

now for the second series:

$\displaystyle \frac{1}{2n+4} > \frac{1}{4n+4} = \left(\frac{1}{4}\right)\left(\frac{1}{n+1}\right)$.

if:

$\displaystyle \sum_{n=1}^\infty \frac{1}{2(n+2)}$ converged, it would follow that:

$\displaystyle \sum_{n=1}^\infty \frac{1}{n+1}$ converged as well. can you see why this can't be true? - Oct 2nd 2012, 11:49 PMMathIsOhSoHardRe: How do you test for convergence or divergence?
I think I would use the comparison test here so we have:

$\displaystyle \frac{1}{2(n+2)}<\frac{1}{n+1}<\frac{1}{n}$

Since we know that $\displaystyle \frac{1}{n}$ is a harmonic series and it is divergent, then according to the comparison test, $\displaystyle \frac{1}{n+1}$ must also be divergent.

Dividing them both we get:

$\displaystyle \frac{\frac{1}{2(n+2)}}{\frac{1}{n+1}}=\frac{1}{2}$

Since $\displaystyle \frac{1}{2}>0$ we know for sure that if one is divergent, then the other must be divergent as well. - Oct 3rd 2012, 11:39 AMDevenoRe: How do you test for convergence or divergence?
*sigh*

if the series $\displaystyle \sum_{n=1}^\infty \frac{1}{n}$ dominates $\displaystyle \sum_{n=1}^\infty \frac{1}{2(n+2)}$ that tells us NOTHING.

what we DO know is that the latter series dominates:

$\displaystyle \left(\frac{1}{4}\right)\left(\sum_{n=1}^\infty \frac{1}{n+1}\right)$

$\displaystyle =\left(\frac{1}{4}\right)\left(\sum_{n=2}^\infty \frac{1}{n}\right)$, which diverges.

basically:

less than a divergent series = ????

more than a convergent series = ????

less than a convergent series = convergent (for non-negative only!)

more than a divergent series = divergent

*********

as to "why", you can think about it this way: if we are adding terms that are reciprocals of polynomials in n, say p(n), linear polynomials don't grow "fast enough" so their reciprocals "shrink fast enough" to stop "adding meat to our infinite sum". but if deg(p) > 1, then the shrinkage is fast enough.

it can be very difficult to tell if 1/f(n) for an increasing function f shrinks fast enough for the series built on it to converge. there are two opposing forces at work:

1. we are adding more and more terms, so the partial sums keep increasing.

2. the terms are getting smaller and smaller, so the rate of increase is decreasing.

so checking for series convergence/divergence isn't an "automatic" process (like for example, differentiation) but more of an art form (like integration). the various "tests" for convergence/divergence only let us tell in some circumstances that some series converge or diverge. in fact, one of the BEST tests is the integral test, but evaluating an integral isn't always easy (some functions aren't integrable in "elementary terms").

why does the integral test work so well? because a series is actually a crude approximation for:

$\displaystyle \int_0^\infty \frac{1}{f(t)}\ dt$

and depending on whether you choose the intervals [n-1,n], or [n,n+1] a series is either a lower sum or an upper sum for said integral (one series has "one more term" than the other, but one term alone makes no difference as to convergence or divergence). - Oct 3rd 2012, 01:15 PMMathIsOhSoHardRe: How do you test for convergence or divergence?
I think I get it now. My mistake was that I used "less than a divergent series" which is inconclusive.

So in order to see if it is divergent, I need a sequence of term that is "more than a divergent series".

So I cannot use $\displaystyle \frac{1}{n}$.

So as you said earlier:

$\displaystyle \frac{1}{2n+4}>\frac{1}{4n+4}$

So we follow the rule that $\displaystyle 0<a_n<b_n$ which makes sure we only work with non-negatives. So if $\displaystyle \sum_{n=1}^{\infty}a_n$ is divergent, then $\displaystyle \sum_{n=1}^{\infty}b_n$ is divergent (more than a divergent series = divergent). If $\displaystyle \sum_{n=1}^{\infty}b_n$ is convergent, then $\displaystyle \sum_{n=1}^{\infty}a_n$ is convergent (less than a convergent series = convergent (for non-negative only!)).

These are the only two rules you can go by. Any other way (which I unfortunately did with $\displaystyle \frac{1}{n}$) is inconclusive and cannot be used.

So basically we have $\displaystyle b_n=\frac{1}{2n+4}$ and we need to find a divergent $\displaystyle a_n$ which is less than $\displaystyle b_n$.

So I'm trying to figure out why you use $\displaystyle \frac{1}{n+1}$.

You write that:

$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n+1}=\sum_{n=2}^{\inft y}\frac{1}{n}$

Now to the right of the equal sign, you write that n=2.

You give the explanation to this here:

Quote:

and depending on whether you choose the intervals [n-1,n], or [n,n+1] a series is either a lower sum or an upper sum for said integral (one series has "one more term" than the other, but one term alone makes no difference as to convergence or divergence)

I hope I understood that correctly.

EDIT: I don't know why a "y" appears on top of the last sum sign. It's supposed to be an infinity sign.