I'll ask the teacher about the greater than and equal to sign, since the greater than sign was what was written on our worksheet. But if it's any comfort, I completely agree
I have another from a tutoring centre and I'm not sure whether it's just a bit advanced for the course or where it fits in exactly into harder 3u (i think it's applications of inequalities)
Considering 1/x (take x>0), explain why 1/n+1 < the integral from (n+1) to n of dx/x < 1/n
As the integral can be simplified to ln(1 +1/x), do we just graph the intervals to prove that it works (I checked on desmos and it fits nicely)
b) hence or otherwise, deduce (1+1/n)^n < e < (1 + 1/n)^(n+1)
Do we separate the two sides from part a) and then raise it to e (e.g. from 1+1/k to e^ (1/k+1)) and multiply by the two sides and somehow manipulate the result? I'm not sure on how to get the lone e
c) this one is an extension question but it goes like this:
((1+n)^n)/(e^n) <n! < ((1+n)^(1+n))/(e^n)
Would that be relevant for this course or not?
Like yeah it's just harder 3U. Since you posted it on the 4U thread that's fine. I'm gonna swap out all the \(n\)'s for \(k\)'s, except for in the last part.
Yeah so for part i), just sketch the right branch of \(y= \frac1x\) and plot the points \( \left(k, \frac1k\right) \) and \(\left( k+1, \frac{1}{k+1} \right)\) on it. Note that the area under \(y = \frac{1}{x}\) from \(k\) to \(k+1\) has area given by \( \int_k^{k+1} \frac{1}{x}\,dx \). Note that a rectangle with length \(1\) and breadth \( \frac{1}{k+1} \) can be drawn right below it, so comparing areas we have \( \frac{1}{k+1} < \int_k^{k+1} \frac{1}{x}\,dx\). The other part is done similarly.
And yep. Upon evaluating, \( \int_k^{k+1} \frac{1}{x}\,dx = \ln \left( 1 + \frac1{k}\right)\), so you can rearrange by chopping the three-sided inequality into 2 two-sided inequalities first.
\[ \text{With your third part, note that rewriting the inequality in part ii) we have}\\ \boxed{ \left(\frac{k+1}{k}\right)^k < e < \left(\frac{k+1}{k}\right)^{k+1} } \]
\[ \text{Now observe that by subbing in }k=1,2,3,\dots, n\text{ we obtain}\\ \left(\frac21\right)^1 < e < \left(\frac21\right)^2\\ \left(\frac32\right)^2 < e < \left(\frac32\right)^2\\ \left( \frac43\right)^2 < e < \left(\frac43\right)^3\\ \vdots\\ \left( \frac{n+1}{n}\right)^n < e < \left(\frac{n+1}{n}\right)^{n+1}\]
\[ \text{Upon multiplying all of that out, we obtain}\\ \boxed{\left(\frac21\right)^1 \left(\frac32\right)^2 \left( \frac43\right)^3 \dots \left( \frac{n+1}{n}\right)^n < e^n < \left( \frac21\right)^2\left(\frac32\right)^3\left(\frac43\right)^4\dots\left(\frac{n+1}n\right)^{n+1}}\]
\[ \text{On the LHS, observe that}\\ 2^1\text{ in the top cancels out with }2^2\text{ in the bottom to yield }\frac12.\\ 3^2\text{ in the top cancels out with }3^3\text{ in the bottom to yield }\frac13\\ \text{and this cancellation process continues. Ultimately}\\ \text{we're just left with }\frac{(n+1)^n}{n!}\text{ after doing all the cancellations.} \]
Note that the \( (n+1)^n\) in the top has nothing in the bottom to be cancelled out with.
\[ \text{An essentially similar thing occurs on the RHS and leaves us }\frac{(n+1)^{n+1}}{n!}\\ \text{So the whole thing boils down to}\\ \boxed{\frac{(n+1)^n}{n!} < e^n < \frac{(n+1)^{n+1}}{n!}}.\]
This immediately rearranges to produce the required result.