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ZarkonBest ResponseYou've already chosen the best response.5
I can already feel my brain shutting down ;)
 2 years ago

bahrom7893Best ResponseYou've already chosen the best response.1
mine shutdown like 3 hours ago
 2 years ago

ZarkonBest ResponseYou've already chosen the best response.5
part b looks like induction
 2 years ago

FoolForMathBest ResponseYou've already chosen the best response.0
part b is definitely induction ... this property could be used to show that 2<e<3
 2 years ago

ZarkonBest ResponseYou've already chosen the best response.5
for n=1 we have 1/1!=1 and 1/2^(11)=1/1=1 \(1\le 1\) so the basis step holds assume \[\frac{1}{k!}\le\frac{1}{2^{k1}}\] is true for some \(k\) then \[\frac{1}{(k+1)!}=\frac{1}{k!}\frac{1}{k+1}\le\frac{1}{2^{k1}}\frac{1}{k+1}\] since \(\frac{1}{2}\ge\frac{1}{n+1}\) for \(n\ge1\) we have \[\frac{1}{2^{k1}}\frac{1}{k+1}\le\frac{1}{2^{k1}}\frac{1}{2}=\frac{1}{2^{k}}\] QED
 2 years ago

FoolForMathBest ResponseYou've already chosen the best response.0
Prof Zarkon I need your inputs on another problem, i have pinged you from the relevant thread. Please do check it :)
 2 years ago

FoolForMathBest ResponseYou've already chosen the best response.0
btw for part c the upper bound is \( 3\).
 2 years ago

bahrom7893Best ResponseYou've already chosen the best response.1
ty guys, i can die a peaceful death now, knowing that 9.3 is finally done...
 2 years ago

FoolForMathBest ResponseYou've already chosen the best response.0
when (exactly) are you going to die? :P
 2 years ago

bahrom7893Best ResponseYou've already chosen the best response.1
not any time soon, hopefully.
 2 years ago
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