Question

Zarkon last one!

Answer 1

I can already feel my brain shutting down ;)

Answer 2

mine shutdown like 3 hours ago

Answer 3

Part A's easy

Answer 4

yes...i get 65/24

Answer 5

1+1+1/2+1/6+1/24

Answer 6

YAY :D

Answer 7

part b looks like induction

Answer 8

part b is definitely induction ... this property could be used to show that 2<e<3

Answer 9

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

Answer 10

Prof Zarkon I need your inputs on another problem, i have pinged you from the relevant thread. Please do check it :)

Answer 11

btw for part c the upper bound is \( 3\).

Answer 12

ty guys, i can die a peaceful death now, knowing that 9.3 is finally done...

Answer 13

when (exactly) are you going to die? :P

Answer 14

not any time soon, hopefully.