Power series: How would you write this off as?

[Archy]

So. There's this question about power series that will eventually take the form of

p= |x| lim _n->inf | nⁿ / (n+1)⁽ⁿ⁺¹⁾ |

But of course, in a futile attempt at a solution I tried doing the derivative for both functions. Didn't get anywhere of course.

Knowing that eventually the answer is x= inf or undefined, I tried to write the equation off as thus:

p= |x| lim _n->inf e^{ln (nn / (n+1)(n+1))}
p= |x| lim _n->inf e^{ln n ln n- n ln (n+1)- ln (n+1)}

Which will of course eventually lead to something along the lines of
p= |x| lim _n->inf e^inf

Which... Effectively writes the whole equation off as p= inf. Convenient but not convincing.

Help here please, thank you.

 

[fzero]

So. There's this question about power series that will eventually take the form of

p= |x| lim _n->inf | nⁿ / (n+1)⁽ⁿ⁺¹⁾ |

But of course, in a futile attempt at a solution I tried doing the derivative for both functions. Didn't get anywhere of course.

Knowing that eventually the answer is x= inf or undefined, I tried to write the equation off as thus:

p= |x| lim _n->inf e^{ln (nn / (n+1)(n+1))}
p= |x| lim _n->inf e^{ln n ln n- n ln (n+1)- ln (n+1)}

Which will of course eventually lead to something along the lines of
p= |x| lim _n->inf e^inf

Which... Effectively writes the whole equation off as p= inf. Convenient but not convincing.

You should be a bit more careful about the limit. The second line above should read

p= |x| lim _n->inf e^{n ln n- n ln (n+1)- ln (n+1)}

As a first step to computing the limit you should compare n \ln n to n \ln(n+1). I don't believe that your stated result is correct. Of course saying anything about the value of |x| would require more information about the value of p than you've given.

 

[Archy]

Sorry about that. Was a typo.
anyway the question requires us to find the range of x in which the power series converges.

original equation is a_n=xⁿ/nⁿ
By the ratio test for absolute convergence,
p= lim_x->inf |a_n+1/a_n|
p= lim_x->inf | [xⁿ⁺¹/ (n+1)⁽ⁿ⁺¹⁾] x [nⁿ/xⁿ] |
p= |x| lim_x->inf nⁿ/(n+1)⁽ⁿ⁺¹⁾

And it continues to it's current state.
The value of p is anonymous and dependent on x.

The answer is that for all values of x, the series diverges.

 

[fzero]

Again, be very careful when computing the limit. I'm not sure where you got the answer, but this series converges for finite x and this ratio test proves it. You can also use a comparison with the series based on b_n = (1/2)^n to find an upper bound of 3/2 for your series evaluated at x=1.

 

[Archy]

Whoops.
Anyway I rang up the lecturer and the answer was actually, p is infinite for all x but the series converges for all values of x.
Sorry again about that.

 

[fzero]

Whoops.
Anyway I rang up the lecturer and the answer was actually, p is infinite for all x but the series converges for all values of x.
Sorry again about that.

What you're saying makes no sense. If p were infinite, the series would diverge for all nonzero x. What happens is that limit is zero, so p is zero for all finite x.