Proof of the Ratio Test and the Triangle Inequality

Fiz2007
Messages
6
Reaction score
0

Homework Statement



Prove: If the limit inf as k goes to infinity of abs(ak+1 / ak) > 1 then the sum from 1 to infinity of ak diverges


Homework Equations




The Attempt at a Solution



So far I have this:

Suppose lim inf abs(ak+1/ak) >1
then, there exists an r such that lim inf abs(ak+1/ak) > r > 1
then \exists N an element of the natural numbers such that k >= N implies
abs(ak+1/ak) > r
that is, for k >= N, abs(ak+1) > abs(ak) r
and, abs (an+1) > r abs(an)
abs (an+2) > r abs (an+1) > r^2 abs(an)
and in general, abs (an+k) > r^k abs(an)

the series the sum from 1 to infinity of abs(an) * r^k diverges to infinity (geometric series with r>1).

Therefore the sum from 1 to infinity of abs(an+k) diverges to infinity by the comparison test.

Then I'm stuck. I don't know how to get from the absolute value diverging to the series diverging. It obvioulsy does not converge absolutely, but what about conditional convergence. Please help!

Thanks
 
Physics news on Phys.org
Fiz2007 said:

Homework Statement



Prove: If the limit inf as k goes to infinity of abs(ak+1 / ak) > 1 then the sum from 1 to infinity of ak diverges


Homework Equations




The Attempt at a Solution



So far I have this:

Suppose lim inf abs(ak+1/ak) >1
then, there exists an r such that lim inf abs(ak+1/ak) > r > 1
then \exists N an element of the natural numbers such that k >= N implies
abs(ak+1/ak) > r
that is, for k >= N, abs(ak+1) > abs(ak) r
and, abs (an+1) > r abs(an)
abs (an+2) > r abs (an+1) > r^2 abs(an)
and in general, abs (an+k) > r^k abs(an)

the series the sum from 1 to infinity of abs(an) * r^k diverges to infinity (geometric series with r>1).

Therefore the sum from 1 to infinity of abs(an+k) diverges to infinity by the comparison test.

Then I'm stuck. I don't know how to get from the absolute value diverging to the series diverging. It obvioulsy does not converge absolutely, but what about conditional convergence. Please help!

Thanks

You've shown the absolute values of the terms in your series are bounded below by a geometric series whose terms go to infinity. The terms in a series have to go to zero if it will have any hope of converging.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

Similar threads

Replies
2
Views
2K
Replies
4
Views
1K
Replies
2
Views
1K
Replies
1
Views
1K
Replies
6
Views
1K
Replies
3
Views
1K
Back
Top