Someone help with this question!!! (1 Viewer)

[forstudents]

Thanx! appreciated if u could help.

 

[Raginsheep]

Easier way:
1/(2p+1) > 1/(2p+2)
Thus
1/(2p+1) + 1/(2p+2) >
1/(2p+2) + 1/(2p+2) =
2/(2p+2)=
1/(p+1)

 

[forstudents]

thancx guys really helpful.thanx

 

[forstudents]

thats ok man i couldn't do most of it u did better than me..lol..Thanx a lot once again!

 

[Yip]

(b) ∫_m^m+1[1/t]dt>1/(m+1)
∫_m+1^m+2[1/t]dt>1/(m+2)
......
∫_2m-1^2m[1/t]dt>1/(2m)

∫_m^m+1[1/t]dt +∫_m+1^m+2[1/t]dt+....+∫_2m-1^2m[1/t]dt=log[(m+1)/m]+log[(m+2)/(m+1)]+....+log[2m/(2m-1)]
=log[2m/m]=log2
Also, as proven earlier,
1/(m+1)+1/(m+2)+...+1/(2m)>=37/60

∫_m^m+1[1/t]dt +∫_m+1^m+2[1/t]dt+....+∫_2m-1^2m[1/t]dt>1/(m+1)+1/(m+2)+...+1/(2m)>=37/60
Thus, log2>37/60

 

[forstudents]

hey lottox i dont get one thing..y did u conclude with LHS is less than RHS? doesn't the question ask LHS>RHS? arent u suppose to reverse it again and thus gettin LHS>RHS as in question?

 

[haque]

yip's already done the graph one but i'll provide my own solutions anyway

a) Let m=3
LHS=1/4 +1/5+1/6=37/60 hence true for m=3, assume true for m=k
1/k+1 +1/k+2 +...... +1/2k>= 37/60 consider m=k+1
1/k+2 +1/k+3 +....1/2k+1/2k+1 +1/2k+2>=37/60-1/k+1 +1/2k+1 +1/2k+2

Now we know that 1/2k+1 +1/2k+2 -1/k+1>0
thus 37/60 + 1/2k+1 +1/2k+2 -1/k+1>37/60
thus 1/k+2 +1/k+3 +...........1/2k+2>=37/60 then do the conclusion etc
b) Consider sum of rectangles and area under graph from x=m to x=2m(area graph>rectangles)
Integral 2m to m 1/t dt>1/m+1 +1/m+2 + .......1/2m>=37/60 (from above)
ln2m-lnm>37/60 hence ln2>37/60