Show sequence eventually strictly increases or eventually strictly decreases. {n/ ln(n)} +infinity, n=2
Can you draw it as is in the question? Use the "draw button".
\[\left\{ \frac{ x }{ \ln x } \right\} ^{+\infty}_{n=2}\]
Using n's in the equation instead of x's. Sorry. Asked to show that sequence is eventually increasing or eventually decreasing and determine smallest value of n
Smallest value of n such that what?
Determine smallest value of n for which the sequence is eventually strictly increasing or decreasing.
Here's the graph. All you need is the point where n/ln begins to exist definitely. https://www.google.com/search?q=n%2F+ln(n&rlz=1C1CHFX_enUS524US524&oq=n%2F+ln(n&aqs=chrome.0.69i57j69i58j69i64l2&sourceid=chrome&ie=UTF-8#safe=off&rlz=1C1CHFX_enUS524US524&q=x/ln+x&spell=1&sa=X&psj=1&ei=bWnvUdajEYzy8ASp1IGADQ&ved=0CCoQBSgA&bav=on.2,or.r_qf.&bvm=bv.49641647%2Cd.eWU%2Cpv.xjs.s.en_US.NyLNrjc7wJY.O&fp=3021d1819c6f28ea&biw=1438&bih=839
Which is where ln(n) = 0, therefore at n = 1.
Do you know how to calculate using to derivative rule?
Yeah.
Let's use x instead of n.
Ignore my previous answer. The graph starts from 2 , so 1 is not included.
I've gotten to \[\frac{ \ln (x) - 1 }{ (\ln (x))^{2}}\] now what do I do?
From that, you get x = e^1 = e
How do I find the smallest value of n from e?
I think you should create a new question and make your prompt clearer. The smallest value is -infinity, but it looks like you're needing another type of answer. Try to create a new question and put more info. This one is already convoluted.
Join our real-time social learning platform and learn together with your friends!