bahrom7893
  • bahrom7893
Zarkon last one!
Mathematics
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bahrom7893
  • bahrom7893
Zarkon last one!
Mathematics
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
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Zarkon
  • Zarkon
I can already feel my brain shutting down ;)
bahrom7893
  • bahrom7893
mine shutdown like 3 hours ago
1 Attachment
bahrom7893
  • bahrom7893
Part A's easy

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Zarkon
  • Zarkon
yes...i get 65/24
bahrom7893
  • bahrom7893
1+1+1/2+1/6+1/24
anonymous
  • anonymous
YAY :D
Zarkon
  • Zarkon
part b looks like induction
anonymous
  • anonymous
part b is definitely induction ... this property could be used to show that 2
Zarkon
  • Zarkon
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
anonymous
  • anonymous
Prof Zarkon I need your inputs on another problem, i have pinged you from the relevant thread. Please do check it :)
anonymous
  • anonymous
btw for part c the upper bound is \( 3\).
bahrom7893
  • bahrom7893
ty guys, i can die a peaceful death now, knowing that 9.3 is finally done...
anonymous
  • anonymous
when (exactly) are you going to die? :P
bahrom7893
  • bahrom7893
not any time soon, hopefully.

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