Questions i cant do. help plz (1 Viewer)

[Bellow]

7c) Prove that for any positive integer m,

[2m+1]/[2m+4] <= sqr.root [(3m-2)/(3m+1)]

Hence deduce that (1/2)(5/8)(7/10).....(2n+1/2n+4) <= 1/sqrroot(3n+1)

also

8bi) Show that

r+1/r-1 = 1 + 2/r + 2/r^2 + 2/r^3 + ..... + 2/r^n + 2/r^(n+1)

ii) show that

sum of r=2 to n (ln(r+1) - ln(r-1)) = ln [n(n+1)/2]

iii) Hence prove by induction that

sum of r=2 to n [ln(1 + 2/r + 2/r^2 + 2/r^3 + ..... + 2/r^n + 2/r^(n+1)] = ln [n(n+1)/2]
for n = 2,3,4,....

can any1 post solutions? thx very much.
is any1 stayin up late btw?

EDIT: sorry made a mistake in question 8biii), should be rite now.

 

[robbo_145]

a good nights sleep is the best study

 

[mojako]

7c
(2m+1)/(2m+4) = 1 - 3/(2m+4) <= 1 - 3/(3m+1) (for n>=3)
<= sqrt(1 - 3/(3m+1)) (since the thing is less than 1)
= sqrt((3m-2)/(3m+1))
then check for n=1 and n=2

(1/2)(5/8)(7/10).....(2n+1/2n+4) <= 1/sqrroot(3n+1)
the left side is just the product of the [2m+1]/[2m+4] thing from m=1 to m=n
the right side is actually the product of sqr.root [(3m-2)/(3m+1)] from m=1 to m=n as well
just calculate what (3m-2)/(3m+1) is for m=1, 2 and 3 and you'll see they cancel out.
the general reason of course is that (3[m+1]-2)/(3[m+1]+1)=(3m+1)/(3[m+1]+1)


i dont get 8i... whats n? also 8iii... dont get what it means
anyway for 8ii,
sum from r=2 to r=n of (ln(r+1) - ln(r-1))
= ln3 - ln1 + ln4 - ln2 + ln5 - ln3 + ln6 - ln4 + ln7 - ln5 + ln8 - ln6 + ...
you can see that the terms cancel out, except -ln1 and -ln2 at the start, and also ln7 and ln8 at the end, which correspond to ln(n) and ln(n+1) when n=7
= -ln1 -ln2 +ln(n) +ln(n+1)
= -ln2 +ln(n) +ln(n+1)
= ln [n(n+1)/2]