Finding the Radius & Interval of Convergence of a Power Series

[craig16]

Homework Statement

\sum from n=1 to inf (1+ 1/2 + ... 1/n)x^n

Find the radius of convergence and the interval of convergence of the given power series.

Homework Equations

Dunno..

The Attempt at a Solution

Stuck thinking about it. I'm not sure if I can combine what's in brackets with the x^n or if I need to do something else.

 

[Dick]

Try a ratio test.

 

[craig16]

Yeah,

Do you think that I should rewrite ( 1 + 1/2 + ... 1/n) as 1/n and do a ration test on
\sum x^n / n

Or is there a special way to treat the first expression in brackets.

 

[craig16]

Hmm, I think you just do the sum of a sum..

But if you do that and use the ratio test you get the limit to be zero, and then it would be convergent for all values of x. I just have no clue if that's how it's done.

 

[Dick]

Yeah,

Do you think that I should rewrite ( 1 + 1/2 + ... 1/n) as 1/n and do a ration test on
\sum x^n / n

Or is there a special way to treat the first expression in brackets.

But that's not true. The ratio test expression is (1+1/2+...+1/n+1/(n+1))/(1+1/2+...+1/n). Can you show that the limit of that is 1?

 

[craig16]

Hmm,

But how would showing the sum in brackets goes to one with the ratio test help? If that limit equals one the ratio test would be inconclusive.

If I treat this as two sums the (1/n) sum would diverge. So I still don't know what to do.

 

[Dick]

Hmm,

But how would showing the sum in brackets goes to one with the ratio test help? If that limit equals one the ratio test would be inconclusive.

If I treat this as two sums the (1/n) sum would diverge. So I still don't know what to do.

I omitted the x^(n+1)/x^n part of the ratio test. If you can show the limit of what I gave you is 1, then the ratio test will give you x for the ratio.

 

[craig16]

Okay, makes sense.

honestly though I don't quite understand the algebra of that limit

n+ 1 / n

you would have, (1/2 + 1/3 ... 1/n+1 ) / (1+1/2... 1/n)

Everything cancels out but ( 1/ n+1 ) / 1 ) ?

Or does the 1/n stay on the bottom

 

[Dick]

Okay, makes sense.

honestly though I don't quite understand the algebra of that limit

n+ 1 / n

you would have, (1/2 + 1/3 ... 1/n+1 ) / (1+1/2... 1/n)

Everything cancels out but ( 1/ n+1 ) / 1 ) ?

Or does the 1/n stay on the bottom

Break it up as (1+1/2+...+1/n)+1/(n+1) in the numerator and (1+1/2+...+1/n) in the denominator. Divide through by the denominator. You get 1+something. What's the something? Can you show the something goes to zero as n->infinity?

 

[craig16]

Okay,

I'm curious though why you wrote the numerator with a 1 in it. Since it is n+1 won't it start at 1/2 and finish with 1/n+1, and the denominator start as 1 and finish with 1/n?

 

[Dick]

Okay,

I'm curious though why you wrote the numerator with a 1 in it. Since it is n+1 won't it start at 1/2 and finish with 1/n+1, and the denominator start as 1 and finish with 1/n?

Not at all. If a_n is the coefficient, then a_1=1, a_2=1+1/2, a_3=1+1/2+1/3. So what's a_3/a_2?