converge or diverge? and how to prove it:
$$\sum_{1}^{inf} (-1)^{n-1}\frac{2^n}{n^4}$$

Question

converge or diverge?  and how to prove it:
$$\sum_{1}^{inf} (-1)^{n-1}\frac{2^n}{n^4}$$

amistre64 · Answer

I had tried, but not confident enough to know if im right :)

$$\lim_{n->inf}\ \sqrt[n]{\left|\frac{2^n}{n^4}\right|}$$
$$\lim_{n->inf}\ \frac{2}{n^{4/n}}$$
$$\frac{2}{n^{0}}=2$$
$$2\cancel\lt1$$
diverges

mr.math · Answer

What do you think about $\large \lim_{n \to \infty} \frac{2^n}{n^4}?$

amistre64 · Answer

isnt that inf/inf?  are we good to lhop it?

mr.math · Answer

I think so!

amistre64 · Answer

2^n (ln(2))^4
-----------   might be the lhop; whic is inf
       4!

mr.math · Answer

If you want to apply L'Hopital"s rule, you will have to do it 4 times in order to get rid of the n in the denominator. For me, it's clear that 2^n grows much faster than n^4 and therefore the limit is infinity.

amistre64 · Answer

yeah, the blurb at the start got me on that one; but then the limit doesnt matter about the start, it only cares about the farout

amistre64 · Answer

mighta been a different blurb on another problem im thinking of :)

mr.math · Answer

I don't get what you mean. Your English is a little bit difficult for me, sorry! :D

amistre64 · Answer

thats ok, i resolved my own quandry :)  thnx

mr.math · Answer

You're welcome :)

mr.math · Answer

Oh and being not able to understand you is my problem, I need to work on my English. Just had to say that!!

anonymous · Answer

$$2^x$$ grows faster than any polynomial in x

zarkon · Answer

you really cant write this  $$\frac{2}{n^{0}}=2$$

if you took the limit there is not n

zarkon · Answer

*should be no n

anonymous · Answer

$$\lim_{n\rightarrow \infty}n^{\frac{4}{n}}=\lim_{n\rightarrow \infty}e^{\frac{4log(n)}{n}}=1$$should work