Series & Minimum/Maximum

Question

Let Vr = (Ur+1)^2 - (Ur)^2 , where Ur+1 & Ur denote the r+1th and rth term of a sequence where Ur>0

It is given that Sigma(from r=1 to n) Vr = [ (n-2)((2n-3)]/ (n^2 + 1) , where n>= 1 and Ur>0

Show that (1/2)<= (Ur+1)^2 - (Ur)^2 <2

My first though was to apply limits, resulting the right side of the equation and then appying Mathematical Induction gave the left hand side of the inequality.

Could you please guide me through a possible method of solving the above simulataneously, or without using principles of Mathematical Induction.

Or what could possibly be the most appropriate method for solving this?.

Thanks!

Answer 1

Lower limit in the expression should be -1/2 (Notice negative sign) . The upper limit can be found by

evaluating limit Vr = (Ur+1)^2 - (Ur)^2 n => infinity. Here it how it is done

For lower limit r=1, n=1 Vr = (2n^2 -5n +2)/(n^2+1) multiply factors using foil in numerator.

Plug n=1 Vr = [2(1)^2 -5(1) +2] /[(1)^2 +1] = -1/2

For upper limit take limit of expression as n=> infinity

Lim (2n^2 -5n + 2)/(n^2+1)

n-> ∞

Lim (2 - 5/n +1/n^2)/(1 + 1/n^2) divide numerator and denominator by n^2

n-> ∞

= 2

Therefore -1/2 ≤ (Ur+1)^2 - (Ur)^2 ≤ 2 Evaluate limit without mathematical induction

I hope this clarifies.