HypeBeast23 said:
Homework Statement
What is the limit as n approaches infinity of: (4n+2^n)/(5n+3^n)
AND
Does the series from 1 to infinity of (1+n)/sqrt(1+n^6) converge absolutely, conditionally or diverge?
Homework Equations
L'hopital's rule, convergence tests...
The Attempt at a Solution
For the limit, I attempted to use l'hopital's rule but was left stuck with a 2^n*ln2 at the top and a 3^n*ln3 at the bottom.
For the series question, I had no idea which test to use :S
Any help would be greatly appreciated!
It shouldn't be hard to see that 2^n and 3^n will be much larger than 4n and 5n so, for very large n, this is essentially \frac{2^n}{3^n}= \left(\frac{2}{3}\right)^n which, since \frac{2}{3}< 1, goes to 0.
For a more rigorous proof, use L'Hopital's rule.
Strictly speaking, L'Hopital's rule only applies to limits of
functions, but since f(x_n), as x_n\to a must converge to \lim_{x\to a} f(x), we can apply it to sequences as well.
Here, the derivative of 4n+ 2^n is 4+ 2^n ln(2) and the derivative of 5n+ 3^n is 4+ 3^nln(3). Since
\frac{4+ 2^n ln(2)}{5+ 3^n ln(3)}
is still "infinity over infinity", apply L'Hopital a second time, getting
\frac{2^n(ln(2))^2}{3^n (n(3))^2}= \frac{ln(2)}{ln(3)}\left(\frac{2}{3}\right)^n
which clearly converges to 0.
As for \sum_{n=0}^\infty (1+n)/\sqrt{1+n^6}, since it involves only positive numbers, there is no question of "conditional convergence"- it either diverges or converges absolutely.
You should be able to use the comparison test: show that this is less than some (possibly large) number times n/\sqrt{n^6}= n/n^3= 1/n^2 and since the series \sum 1/n^2 converges (by the integral test), this series converges absolutely.
