No error! Look at my fourth step about half way in! I typoed a minus to a plus in the natural log, its my fault, I do these things on the computer not on pencil and paper, so when I make a typo it carries through the rest of the calculation
So yes you are right, my work will have a negative sign differenece
Good job Paul, here was my solution
Consider a function
Now let
So now we must find where goes as goes to zero
Which means we must find
Now by letting
we get
Which by the realtion of the Ratio test and Root test is equivalent to
Giving us
And now seeing that this limit falls under the reign of
The limit is one
i think you're assuming that m is an integer. the idea of my solution is very simple but it requires
some writing ... so i wouldn't be surprised if some typos were found in it! ok, first see that:
now, in your integral, let then:
thus by
now since and , we'll have:
and:
it's also clear, by Riemann sum, that:
and:
thus from (2), (3), (4), (5), and (6) we have:
Here is another one I came up with, find L.
Where
Restrictions: You may not use asymptotic equivalences, rate of increase, or in any other way disregard terms.
EDIT: typo, also assuming there is no domain conflictions with other functions