Limit and Convergence Questions

[HypeBeast23]

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!

 

[HallsofIvy]

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.