## bahrom7893 4 years ago Zarkon last one!

1. Zarkon

I can already feel my brain shutting down ;)

2. bahrom7893

mine shutdown like 3 hours ago

3. bahrom7893

Part A's easy

4. Zarkon

yes...i get 65/24

5. bahrom7893

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

6. anonymous

YAY :D

7. Zarkon

part b looks like induction

8. anonymous

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

9. 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

10. anonymous

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

11. anonymous

btw for part c the upper bound is $$3$$.

12. bahrom7893

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

13. anonymous

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

14. bahrom7893

not any time soon, hopefully.