bahrom7893
Zarkon last one!
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Zarkon
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I can already feel my brain shutting down ;)
bahrom7893
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mine shutdown like 3 hours ago
bahrom7893
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Part A's easy
Zarkon
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yes...i get 65/24
bahrom7893
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1+1+1/2+1/6+1/24
Pippa
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YAY :D
Zarkon
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part b looks like induction
FoolForMath
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part b is definitely induction ... this property could be used to show that 2<e<3
Zarkon
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for n=1 we have 1/1!=1 and 1/2^(1-1)=1/1=1
\(1\le 1\) so the basis step holds
assume \[\frac{1}{k!}\le\frac{1}{2^{k-1}}\] is true for some \(k\)
then \[\frac{1}{(k+1)!}=\frac{1}{k!}\frac{1}{k+1}\le\frac{1}{2^{k-1}}\frac{1}{k+1}\]
since \(\frac{1}{2}\ge\frac{1}{n+1}\) for \(n\ge1\) we have
\[\frac{1}{2^{k-1}}\frac{1}{k+1}\le\frac{1}{2^{k-1}}\frac{1}{2}=\frac{1}{2^{k}}\]
QED
FoolForMath
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Prof Zarkon I need your inputs on another problem, i have pinged you from the relevant thread. Please do check it :)
FoolForMath
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btw for part c the upper bound is \( 3\).
bahrom7893
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ty guys, i can die a peaceful death now, knowing that 9.3 is finally done...
FoolForMath
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when (exactly) are you going to die? :P
bahrom7893
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not any time soon, hopefully.