Verify the convergence or divergence of a power series

[DottZakapa]

Homework Statement

$\sum_{k=0}^\infty (-1)^\left(k+1\right) \frac {k} {log(k+1)} (2x-1)^k$

Relevant Equations

convergence divergence tests

At the exam i had this power series

but couldn't solve it

$\sum_{k=0}^\infty (-1)^\left(k+1\right) \frac {k} {log(k+1)} (2x-1)^k$

i did apply the ratio test (lets put aside for the moment (2x-1)^k ) to the series $\sum_{k=0}^\infty \frac {k} {log(k+1)}$ in order to see to what this limit tends.

$\lim_{n \rightarrow +\infty} \frac {k(1+\frac 1 k)} {log k+log(1+\frac 2 k) } \frac {log k +log (1+\frac 1 k)}{k}=\lim_{n \rightarrow +\infty}\frac {(1+\frac 1 k)} {log k+log(1+\frac 2 k) } \left(log k +log (1+\frac 1 k)\right)$

but from here i don't know how to proceed.

 

[member 587159]

Just an idea:

$$\log a - \log b =\log(a/b)$$

 

[DottZakapa]

Just an idea:

$$\log a - \log b =\log(a/b)$$

i don't see where to apply it, there are just additions

 

[Mark44]

Homework Statement:: $\sum_{k=0}^\infty (-1)^\left(k+1\right) \frac {k} {log(k+1)} (2x-1)^k$
Relevant Equations:: convergence divergence tests

At the exam i had this power series

but couldn't solve it

$\sum_{k=0}^\infty (-1)^\left(k+1\right) \frac {k} {log(k+1)} (2x-1)^k$

i did apply the ratio test (lets put aside for the moment (2x-1)^k ) to the series $\sum_{k=0}^\infty \frac {k} {log(k+1)}$ in order to see to what this limit tends.

$\lim_{n \rightarrow +\infty} \frac {k(1+\frac 1 k)} {log k+log(1+\frac 2 k) } \frac {log k +log (1+\frac 1 k)}{k}=\lim_{n \rightarrow +\infty}\frac {(1+\frac 1 k)} {log k+log(1+\frac 2 k) } \left(log k +log (1+\frac 1 k)\right)$

but from here i don't know how to proceed.

Surely the series converges for x = 1/2, so the problem boils down to determining whether the interval of convergence is just the point x = 1/2, or is some larger interval centered on 1/2.

Nit: Your limits should be on k, not n.

When you use the Ratio Test, you should end up with $|2x - 1|\lim_{k \to \infty}\frac{(k + 1)\log(k + 1)}{k\log(k + 2)}$. I haven't worked this through, but I would see if L'Hopital's Rule could be of use here.

 

[LCKurtz]

It is pretty easy to show that both $\frac{k+1}{k}$ and $\frac{\log(k+2)}{\log(k+1)}$ go to $1$.

 

[DottZakapa]

It is pretty easy to show that both $\frac{k+1}{k}$ and $\frac{\log(k+2)}{\log(k+1)}$ go to $1$.

you right, so the radius of convergence become (0,1)

 

[Mark44]

you right, so the radius of convergence become (0,1)

That would be the interval of convergence, not the radius. (I haven't worked the problem, so can't confirm that your result is correct.)
If your result is correct, the radius of convergence would be R = 1/2.

Don't forget to check the endpoints of the interval.

 

[DottZakapa]

That would be the interval of convergence, not the radius. (I haven't worked the problem, so can't confirm that your result is correct.)
If your result is correct, the radius of convergence would be R = 1/2.

Don't forget to check the endpoints of the interval.

If

It is pretty easy to show that both $\frac{k+1}{k}$ and $\frac{\log(k+2)}{\log(k+1)}$ go to $1$.

Then R =1

Which implies

|2x-1|<1

No?

 

[Mark44]

Edited to correct my error:
Right. So the radius of convergence is 1/2. Solve your inequality to get the interval of absolute convergence. Finally, if the problem asks for it, determine whether the series converges conditionally at each endpoint of the convergence interval.

 

[LCKurtz]

Right. So the radius of convergence is 1.

Obviously a typo, he means 1/2.

 

[Mark44]

Obviously a typo, he means 1/2.

Thanks. I was confusing the length of the convergence interval with its radius.

 

[DottZakapa]

Ok, i am lost... from where that 1/2 comes from if the limit goes to 1

 

[Mark44]

Ok, i am lost... from where that 1/2 comes from if the limit goes to 1

Yes, you are lost. You are getting confused on what you're trying to do. The first thing you were doing was to determine whether the series converged absolutely. To do that, you needed to show that $\lim_{n \to \infty} \frac{a_{n+1}}{a_n} < 1$, which occurs if |2x - 1| < 1, or equivalently, 0 < x < 1.

Post #6:

so the radius of convergence become (0,1)

As already noted in post #7, this was the interval of convergence in which the series converged absolutely. The radius of convergence is the distance from the middle of the interval to either end.
Post #7:

If your result is correct, the radius of convergence would be R = 1/2.

Note that the $(2x - 1)^k$ factor in your series could also be written as $2^k(x - 1/2)^k$. So your power series can be thought of as powers of x - 1/2.

 

[LCKurtz]

Ok, i am lost... from where that 1/2 comes from if the limit goes to 1

The ratio test gave you $|2x - 1| < 1$. Factor out a $2$ and divide both sides by it giving $|x - \frac 1 2| < \frac 1 2$. That tells you the center of the series is $\frac 1 2$ and the radius of convergence is $\frac 1 2$.

 

[DottZakapa]

Yes, you are lost. You are getting confused on what you're trying to do. The first thing you were doing was to determine whether the series converged absolutely. To do that, you needed to show that $\lim_{n \to \infty} \frac{a_{n+1}}{a_n} < 1$, which occurs if |2x - 1| < 1, or equivalently, 0 < x < 1.

ok but the limit is 1 so you cannot conclude anything

 

[Mark44]

ok but the limit is 1 so you cannot conclude anything

You can conclude that the limit is less than 1 if |2x - 1| < 1. IOW, your series converges absolutely by the Ratio Test, if |2x - 1| < 1.

 

[LCKurtz]

ok but the limit is 1 so you cannot conclude anything

No. The limit wasn't $1$. It was $|2x-1|\cdot 1$.

 

[DottZakapa]

ok thanks :)