Range of validity

How do i find the range of validity of this series expansion, without making a substitution into the original ln(1+x)?

THanks!

Rewrite your expression on the left in terms of two "ln" functions, and it should then be clear.

Ah right, would it be -1<x<1?

But when i made a substitution of 2x/(1-x) into -1<x<1 which is the range of valid x values for ln(1+x) i got -1<x<1/3?

I think the ratio test works here, but this is the first time I'll be using it so please correct me if I'm wrong.

The nth term of the series on the right hand side is $a_n=\frac{2}{2n-1} x^{2n-1}$, so $L=\lim_{n\rightarrow \infty} \left | \frac{a_{n+1}}{a_n}\right | = \lim_{n\rightarrow \infty} \left | \frac{\frac{2}{2n+1} x^{2n+1}}{\frac{2}{2n-1} x^{2n-1}}\right | = x^2\lim_{n\rightarrow \infty} \left |\frac{2n-1}{2n+1}\right |$

$=x^2\lim_{n\rightarrow \infty} \left |1-\frac{2}{2n+1} \right| = x^2$

Now by the ratio test, for the series to converge we need $L<1$, and since we have shown that $L=x^2$ we must have $x^2<1$ i.e. $-1<x<1$.

Is this right? Anyone?

What does the bit in bold mean?

(Have you asked this question before - I'm sure I corrected someone who was working on this problem a couple of months ago?!)

I dont know if this is the correct approach but.
From data given the sets of x values that are valid for ln(1+x) are -1<x<1
I let 1+y=1+x / 1-x

y=2x/1-x

Substituted this into ln(1+x) to get ln(1+x/1-x) and also the sets of valid x values -1<2x/1-x<1

Which gives -1<x<1/3 i think.

And it wasn't me *shaggy voice.

OK, you're making the same mistake as the earlier poster that I corrected on this.

The range -1 < y < 1 lets you expand ln(1 + y) in powers of y.

So if you put 1 + y = (1+x)/(1-x) you have an expansion that is valid in powers of (1+x)/(1-x) NOT an expansion that is valid in powers of x. Basically you're answering a completely different question

I guess you could write $1+k=\frac{1+x}{1-x}$, solve for k in terms of x (i.e. just subtract 1 and manipulate right hand side a bit), and then solve -1<k<1 ?

So why does it work for ln(1+2x)?
Subbing 2x in for x in ln(1+x)?

Which gives valid x values of -1/2<x<1/2?

ThankS!

I guess you could write $1+k=\frac{1+x}{1-x}$, solve for k in terms of x (i.e. just subtract 1 and manipulate right hand side a bit), and then solve -1<k<1 ?

Ah, yes. Just read some of the previous posts.

I'm still not sure why it gives -1<x<1/3. Man I'm bad at calculus.

Because you expand in powers of (2x) you get some powers of 2 in front of powers of x, i.e. you still really have an expansion in powers of x, but with different numbers in front.



I think that's what he's tried to do, which is why he's got the wrong answer!

The point is, if you have an expansion like

ln(1 + y) = A + By + Cy^2 + ... for -1 < y < 1

then you can put y = 2x to get ln(1 + 2x) = A + B(2x) + (2x)^2 + ... for -1 < 2x < 1

which is still a series in powers of x

BUT if you put something like y = 2x/(1 - x) then you get

ln(1 + y) = ln(1 +(2x/(1-x))) = ln((1+x)/(1-x)) = A + B(2x/(1-x)) + C((2x/(1-x))^2) + ... for -1 < 2x/(1-x) < 1

which isn't what you want at all, because the series on the right isn't in powers of x - it's in powers of 2x/(1-x)!!

That's why I said the OP is answering a different question from finding a series in powers of x

IF YOU SAID THAT THE FIRST TIME I WOULD OF UNDERSTOOD! -.-

Thanks!

Now by the ratio test, for the series to converge we need $L<1$, and since we have shown that $L=x^2$ we must have $x^2<1$ i.e. $-1<x<1$.

Is this right? Anyone?

Thanks. For $x=\pm 1$ we have the series 2+2/3+2/5+... and -2-2/3-2/5-..., which are both divergent since 1+1/3+1/5+... is divergent.

I think this looks fine, except for the question of convergence when $x=\pm1$, which you need to examine separately - the ratio test is inconclusive in that case.

For the ratio test, dont you have to show that the lim as n-> infinity is <1?
For the series to converge?

Thanks!

If that limit < 1, then the series converges.

However, if the limit = 1, then the series may or may not converge; there are convergent series where the ratio test gives 1, and divergent series where the ratio test gives 1. So you can make no conclusion about the convergence of a series in this case. You need to use other techniques in this case.

So when arkanm set $x^2 < 1$, he could find a range where the series definitely converged, but he could make no statement about its convergence when $x =1$. He had to do that separately.

He had no need to exam the case where $x^2>1$ since the ratio guarantees divergence then.

(In fact, the ratio test guarantees a stronger form of convergence, called absolute convergence, when the limit < 1)

Ahh right thanks, how does the ratio test actually work?

At this stage you really need to be looking at a resource like Wikipedia, or grabbing yourself an analysis book to look at - a proper proof of the ratio test would basically involve somebody copying out a chunk of analysis proof for you!

(It's not difficult to prove, but it's not really a good use of someone else's time getting them to type out manually what you can probably find online or in a book that deals with series rigorously)

Are you an A level student btw - you seem to be posting questions across a wide range of topics, but I'm not clear what qualification you're studying for?