Question

This has to do with sequences.
Find the limit of....

Answer 1

\[\lim_{n \rightarrow } \frac{ (-1)^{\frac{ n(n+1) }{ 2 }}\ln n }{ n ^{2} }\]

Answer 2

n is to infinity

Answer 3

The way I was always taught was to look and see which terms to go infinity, and how quickly they do it. So as n goes to infinity what does
\[ -1^{n(n+1))} \]
do?

Answer 4

does it get bigger or smaller or stay the same?

Answer 5

How about
\[ \ln n \]
? Doe it get bigger, smaller, or stay the same?



How about \[ \frac{1}{n^2} \]?

Answer 6

okay, so after I find out whether it gets larger, smaller, or stays the same, what do I do next?

Answer 7

Sorry about the wait. Technical difficulties :P

Anyways, you compare them. If the denominator goes to infinity "faster" than the numerator the limit goes to zero. If the numerator is "faster," it goes to infinity. If you can't easily see it then it might converge at a value. We're lucky in this instance, however!

logarithms go to infinity much "slower" than exponentials, easily seen by showing that if n is 100,

ln 100 ~ 4.6
100^2 = 10000

So, since the first term is just always 1, an (ln x)/x^2 is heavily weighted on the denominator, ultimately the limit of our function goes to zero as x goes to infinity. Does that make sense? It's not as rigorous as it is holistic, but is a good way of picturing what's happening in the function...

Answer 8

okay, thank you. Now I understand what my professor was talking about during lecture.