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

bahrom7893
 3 years ago
Best ResponseYou've already chosen the best response.2mine shutdown like 3 hours ago

Zarkon
 3 years ago
Best ResponseYou've already chosen the best response.5part b looks like induction

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

Zarkon
 3 years ago
Best ResponseYou've already chosen the best response.5for 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

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

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

bahrom7893
 3 years ago
Best ResponseYou've already chosen the best response.2ty guys, i can die a peaceful death now, knowing that 9.3 is finally done...

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

bahrom7893
 3 years ago
Best ResponseYou've already chosen the best response.2not any time soon, hopefully.
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