Finding the Interval of Convergence for the Taylor Series of ln(x) at a=7

[lxman]

Homework Statement

The Taylor series of function f(x)=ln(x) at a=7 is given by:

f(x)=\sum^{\infty}_{n=0}c_{n}(x-7)^{n}

Determine the interval of convergence

The Attempt at a Solution

I have worked out that the series would be of the form:

ln(7)+\frac{x-7}{7}-\frac{(x-7)^{2}}{(2)7^{2}}+\frac{(x-7)^{3}}{(3)7^{3}}...

Am I ok to this point?

 

[micromass]

Yes, this looks good!

 

[vela]

Looks good so far.

 

[lxman]

Thank you.

Now, in order to find the interval of divergence, as I understand it, I have to apply the ratio test. In order to do this, I have to write the above sequence in sigma notation. I have arrived at:

\sum^{\infty}_{n=1}\frac{(x-7)^{n}(-1)^{n+1}}{n(7)^{n}}

This accounts for all the terms with the exception of the ln(7) on the front. What do I do about that?

 

[micromass]

Thank you.

Now, in order to find the interval of divergence, as I understand it, I have to apply the ratio test. In order to do this, I have to write the above sequence in sigma notation. I have arrived at:

\sum^{\infty}_{n=1}\frac{(x-7)^{n}(-1)^{n+1}}{n(7)^{n}}

This accounts for all the terms with the exception of the ln(7) on the front. What do I do about that?

Just ignore it, the ln(7) won't change the convergence anyway. If the above sum converges, then also

ln(7)+\sum^{+\infty}_{n=1}\frac{(x-7)^{n}(-1)^{n+1}}{n(7)^{n}}

will converge. Leaving out a finite number of terms is completely harmless...

 

[lxman]

Okay, step by step . . . Oh, and btw, I should have said convergence instead of divergence above.

Anyway . . .

I apply the ratio test as such:

\stackrel{lim}{n\rightarrow\infty}\left|\frac{(x-7)^{n+1}}{(n+1)(7)^{n+1}}\ \frac{n(7)^{n}}{(x-7)^{n}}\right|=\stackrel{lim}{n\rightarrow\infty}\left|\frac{n(x-7)}{(n+1)7}\right|

Looking at this, it appears that, whatever value I choose for x, as n increases to \infty, the limit will become one. Does this mean that the interval of convergence is zero?

P.S. What's the proper latex for lim n->oo?

 

[micromass]

Okay, step by step . . . Oh, and btw, I should have said convergence instead of divergence above.

Anyway . . .

I apply the ratio test as such:

\stackrel{lim}{n\rightarrow\infty}\left|\frac{(x-7)^{n+1}}{(n+1)(7)^{n+1}}\ \frac{n(7)^{n}}{(x-7)^{n}}\right|=\stackrel{lim}{n\rightarrow\infty}\left|\frac{n(x-7)}{(n+1)7}\right|

OK, this looks fine.

Looking at this, it appears that, whatever value I choose for x, as n increases to \infty, the limit will become one. Does this mean that the interval of convergence is zero?

I don't quite see why the limit should be 1... Certainly, \lim_{n\rightarrow+\infty}{\frac{2n}{n+1}} is 2 and not 1...

P.S. What's the proper latex for lim n->oo?

It is \lim_{n\rightarrow +\infty} there is no need to use stackrel here...

 

[lxman]

Okay, yes, I understand. I drew an incorrect conclusion there.

Therefore, essentially, my key inequality would boil down to:

|x-7|<1

which means that I end up with:

6<x<8

The endpoints 6 and 8 still need to be tested, though. Going back to my original definition of the series and plugging in 6, I get:

\sum^{+\infty}_{n=1}\frac{(-1)^{n}(-1)^{n+1}}{n(7)^{n}}

=\sum^{+\infty}_{n=1}\frac{1}{n(7)^{n}}

Which series would converge by being a p series with p > 1 (?)

Therefore my interval becomes [6,8

Plugging in 8, I get:

\sum^{+\infty}_{n=1}\frac{1^{n}(-1)^{n+1}}{n(7)^{n}}

=\sum^{+\infty}_{n=1}\frac{-1}{n(7)^{n}}

Which, due to the same reasoning would also converge (?)

If I am correct to this point, my interval of convergence would then be [6, 8]

 

[Dick]

Okay, yes, I understand. I drew an incorrect conclusion there.

Therefore, essentially, my key inequality would boil down to:

|x-7|<1

which means that I end up with:

6<x<8

The endpoints 6 and 8 still need to be tested, though. Going back to my original definition of the series and plugging in 6, I get:

\sum^{+\infty}_{n=1}\frac{(-1)^{n}(-1)^{n+1}}{n(7)^{n}}

=\sum^{+\infty}_{n=1}\frac{1}{n(7)^{n}}

Which series would converge by being a p series with p > 1 (?)

Therefore my interval becomes [6,8

Plugging in 8, I get:

\sum^{+\infty}_{n=1}\frac{1^{n}(-1)^{n+1}}{n(7)^{n}}

=\sum^{+\infty}_{n=1}\frac{-1}{n(7)^{n}}

Which, due to the same reasoning would also converge (?)

If I am correct to this point, my interval of convergence would then be [6, 8]

I think your key inequality is |(x-7)/7|<1 isn't it? What happened to the 7 in the denominator?

 

[lxman]

Yes, I see.

Let me try again. I need:

\left|\frac{x-7}{7}\right|&lt;1

which means that 0<x<14.

So, substituting those values back in, I get:

\sum^{+\infty}_{n=1}\frac{(-7)^{n}(-1)^{n+1}}{n(7)^{n}}

=\sum^{+\infty}_{n=1}\frac{-(-1)^{n+1}}{n}

Hmm . . . so I am left with a series comparable to 1/n, and as 1/n diverges, then this one will also (?)
Or, since this is an alternating series which is decreasing to 0, it would converge (?)

Plugging in 14:

\sum^{+\infty}_{n=1}\frac{(7)^{n}(-1)^{n+1}}{n(7)^{n}}

\sum^{+\infty}_{n=1}\frac{(-1)^{n+1}}{n}

Same questions as above (?)

 

[Dick]

Yes, I see.

Let me try again. I need:

\left|\frac{x-7}{7}\right|&lt;1

which means that 0<x<14.

So, substituting those values back in, I get:

\sum^{+\infty}_{n=1}\frac{(-7)^{n}(-1)^{n+1}}{n(7)^{n}}

=\sum^{+\infty}_{n=1}\frac{-(-1)^{n+1}}{n}

Hmm . . . so I am left with a series comparable to 1/n, and as 1/n diverges, then this one will also (?)
Or, since this is an alternating series which is decreasing to 0, it would converge (?)

Plugging in 14:

\sum^{+\infty}_{n=1}\frac{(7)^{n}(-1)^{n+1}}{n(7)^{n}}

\sum^{+\infty}_{n=1}\frac{(-1)^{n+1}}{n}

Same questions as above (?)

You are being sloppy with signs. If you be a bit more careful, you'll find one of those series doesn't alternate. Which one is it? But yes, the alternating one converges. The other one doesn't.

 

[lxman]

Okay, almost there . . .

I am presuming that:

\frac{(-7)^{n}}{7^{n}}\neq-1

But rather:

=(-1)^{n}

This would then give me:

\frac{1}{n}

which I definitely know diverges. Therefore my interval would be (0, 14]

Have I finally got it right?

 

[Dick]

Yes, I think you have it right. Though that would give you -1/n, yes? The difference isn't that important. It still diverges.

 

[lxman]

Well, the difference is technical, but technicalities can make a difference. :)

I would end up with:

(-1)^{n}(-1)^{n+1}

Since I have common bases - oh, I see - I would just add the exponents, not multiply the bases. So, yes, it would be:

(-1)^{2n+1}

Since 2n+1 will always be an odd power, I effectively have -1 on the top.

Thanks all for the help.