Convergence of an infinite series

[Dollydaggerxo]

Homework Statement
Well I am analysing the convergence of the following series:
$$\sum$$$$\frac{2n}{n^{3}+1}x^{n}$$

from n=0 to infinity.


The attempt at a solution
I have begun by using the ratio test, but as i have the limit in terms of x and n, i can't tell if it is bigger than 0? So does this mean I cannot use the ratio test?
The limit i have as n approaches infinity is:

$$\frac{x(n+1)(1+n^{3})}{n(1+(n+1)^{3}}$$

Any help would be greatly appreciated.

 

[Mark44]

You're missing a right parenthesis in your limit expression. It should be
$$\frac{x(n+1)(1+n^{3})}{n(1+(n+1)^{3})}$$

Expand all of the expressions involving n in the numerator and denominator, and factor out n^4 from each. Since your limit is on n, x doesn't play a role, and can be brought out of the limit expression. In this problem, the limit of the ratio in the ratio test will depend on the value of x. What are the limitations on the values of x so that the limit of the absolute value of the ratio is less than 1?

 

[Dollydaggerxo]

thanks for replying!

well I have expanded it out, but then I get

$$\frac{x(n+1)^2}{n(n+2)}$$

which i don't think seems right?
And if it is, where would i go from here? i don't have any limits for x, does this mean i cannot use the ratio test?

Thanks!

 

[LCKurtz]

If the x wasn't there, what would the limit be? What values of x make the absolute value of the limit less than one?

 

[Dollydaggerxo]

Going back to my first post, I have the limit of the second bit, ignoring the x, to be 0. So if that limit is zero, does that mean the limit of the whole thing is also zero?


I have the limit of xⁿ to be $$\frac{1}{1-x}$$ so that part converges when the absolute value of x is less than one and diverges when the absolute value of x is larger than 1.

I just don't understand how to put them together? am i right to do it separately or am i going the wrong way about it?

any help would be greatly apprectiated!

 

[lanedance]

i think you missed it...

as Mark & LCKurtz hinted consider the limit L (if it converges) of the following expressino
$$lim_{n\to\infty }\frac{(n+1)^2}{n(n+2)}= L$$

so so for your ratio you get
$$lim_{n\to\infty } \frac{x(n+1)^2}{n(n+2)}= xL$$

so what constraint does this put on x for the series to converge? also think about the physical behaviour of the term x^n for differnt x's...

 

[Dollydaggerxo]

right i see,
so the limit for the first expression would be 1,
therefore the limit for the second equation would be x?

and so
if x < 1, then the series converges
if x > 1, then it diverges

but what about if x = 1?
those constraints i used were that of the ratio test, as i had used the ratio test before. so what do i do if x is 1?

I very much appreciate your reply, thank you!

 

[lanedance]

pow!

if x is 1, the ratio test is inconclusive, so you need a different test, I would suggest a comparison or integral test might work well...

 

[Dollydaggerxo]

thank you for your help!

so now I know for x > 1 and x <1, can i just take x to be 1 for my next test?

so for example,

i would do a comparison test on
$$\frac{2n}{n^3+1}1^n$$

or do i have to keep the x?

and if i can replace the x by 1, then 1^n is always going to be 1, so I can ignore that part? or i have i gone way off track here! haha

 

[lanedance]

you can and should do exactly as you have, you now know the series diverges for x>1 & converges for x<1

now do the comparison test with x =1, noting 1^n = 1

 

[Dick]

you can and should do exactly as you have, you now know the series diverges for x>1 & converges for x<1

now do the comparison test with x =1, noting 1^n = 1

Are you guys forgetting the absolute value in the ratio test? You know the series converges for |x|<1 and diverges for |x|>1. There's another value of x to check besides x=1.

 

[Dollydaggerxo]

so i can compare with 1/x^2 and say that it is absolutely convergent and has a limit 0?
do you think i need to state what the limits are in each case?
the question itself just says consider the convergence of the series.

and what would the other value be, Dick?
negative numbers? where do they fit in? because if it was a negative number the series would be alternating wouldn't it? and therefore be convergent? that is what i don't get about the absolute values of x!

Thanks

 

[lanedance]

good point, I jumped to using x only in the range $[0,\infty)$

As Dick points out you should really consider x across the whole real line $(-\infty,\infty)$,

the ratio test gives a interval of convergence for your series |x|< 1, leaving two indeterminant points in the ratio test that you need to check at the edges

 

[Dollydaggerxo]

would the other point be if x was negative.
and am i right in saying that this would make it an alternating series and therefore be convergent?

Thanks guys!

 

[Dick]

would the other point be if x was negative.
and am i right in saying that this would make it an alternating series and therefore be convergent?

Thanks guys!

Yes, the two points to check are x=1 and x=(-1). And yes, you get an alternating series at x=(-1). But not all alternating series converge, what has to be true of the terms? And when you apply a comparion test at x=1, you can't compare with 1/x^2. 1/x^2 isn't a series. Try and keep x and n separate, ok?

 

[Dollydaggerxo]

Yes, the two points to check are x=1 and x=(-1). And yes, you get an alternating series at x=(-1). But not all alternating series converge, what has to be true of the terms? And when you apply a comparion test at x=1, you can't compare with 1/x^2. 1/x^2 isn't a series. Try and keep x and n separate, ok?

Okay, do alternating series converge if $$U_{n} \rightarrow 0$$?
which I think this particular series does.

and sorry I think I meant $$\frac{1}{n^2}$$, would that work?
Thanks again.

 

[Dick]

Okay, do alternating series converge if $$U_{n} \rightarrow 0$$?
which I think this particular series does.

and sorry I think I meant $$\frac{1}{n^2}$$, would that work?
Thanks again.

You are being kind of sloppy here. No. 2n/(n^3+1) isn't less than 1/n^2. Something very much like 1/n^2 will work. For alternating series the absolute value of the terms has to decrease MONOTONICALLY to zero. No idea what you mean by U_n.

 

[Dollydaggerxo]

Thanks for the reply, and I do apologise if I'm not clear, I'm not very clued up on this subject!
I appreciate your patience with me though.

and i know this sounds silly but how do you know when it is less that 1/n^2 or 1/n etc. Do you plug in some terms and check or are we talking about limits being bigger or the sum being bigger?

I have found the limit to be zero anyway, by the comparison test with 1/n. But like i said, not really sure this works.

 

[Dick]

The easiest way to handle a problem like this is to observe (as you did) that 2n/(n^3+1) is sort of like 1/n^2 so it converges and you want to bound from above. Now do something to 2n/(n^3+1) to make a series that is simpler and larger. You can make it larger by making the denominator smaller. So change (n^3+1) to n^3.

 

[Dollydaggerxo]

Right I see, so...

For all n > 0, (does it matter that I'm saying bigger than 0 even though it should start at n=0?)

$$\frac{2n}{n^3+1}<\frac{2n}{n^3}=\frac{2}{n^2}$$

Therefore, by comparison with $$\frac{1}{n^2}, \frac{2n}{n^3}$$ converges.

Therefore, $$\frac{2n}{n^3+1}$$ should also converge because by the comparison test, if the 'larger' series converges, then so does the 'smaller' series. (For non-negative series!)

I wasn't sure what to do about the n=0 thing, i can't have a term starting at zero because it would make the denominator 0 for the $$\frac{2n}{n^3}$$

Any help would be appreciated :)

 

[Dollydaggerxo]

Also, For the second part, with x=-1 I have found that

$$\sum(-1)^n\frac{n}{n^2+1}$$

converges.

I have done this using alternating series test, can somebody just confirm if this is correct please.
Thankyou.

 

[lanedance]

yes that series converges, all you need to show for the alternating series test is that the magnitude of terms tend monotonically to zero for large n

 

[Dollydaggerxo]

brillian thank you very much!
it is just the other bit I am still a bit confused on now then.

 

[Dick]

Right I see, so...

For all n > 0, (does it matter that I'm saying bigger than 0 even though it should start at n=0?)

$$\frac{2n}{n^3+1}<\frac{2n}{n^3}=\frac{2}{n^2}$$

Therefore, by comparison with $$\frac{1}{n^2}, \frac{2n}{n^3}$$ converges.

Therefore, $$\frac{2n}{n^3+1}$$ should also converge because by the comparison test, if the 'larger' series converges, then so does the 'smaller' series. (For non-negative series!)

I wasn't sure what to do about the n=0 thing, i can't have a term starting at zero because it would make the denominator 0 for the $$\frac{2n}{n^3}$$

Any help would be appreciated :)

When you are doing a comparison test (or a lot of other tests for that matter) the first few terms of the series don't matter. In fact you can drop any finite number of terms from the comparison. All that really matters is that the comparison has to hold for all n sufficiently large. You are just trying to show convergence of the series, not actually estimate it's size.

 

[Dollydaggerxo]

ahh i see, that makes a lot more sense now! thanks.
So is it immediately true that if the larger series converges, so does the smaller? I have found a similar question on another site and it tells me that it diverges so I am a bit confused.

Can you please confirm if $$\frac{2n}{(n^3+1)}$$ is convergent so I can work 'backwards' to prove it as I am getting far too many answers?

Thanks

 

[Dick]

ahh i see, that makes a lot more sense now! thanks.
So is it immediately true that if the larger series converges, so does the smaller? I have found a similar question on another site and it tells me that it diverges so I am a bit confused.

Can you please confirm if $$\frac{2n}{(n^3+1)}$$ is convergent so I can work 'backwards' to prove it as I am getting far too many answers?

Thanks

Yes, it converges. It's less than 2/n^2. 2/n^2 converges by the integral test. Or just by noticing it's twice the p-series 1/n^2.

 

[Dollydaggerxo]

Okay, I think I have got the jist of it now. Thank you so much for all your help! both of you!