Solve the inequality

Question

(k+1)/ 2^k <= 1/8

 

please help 

Answer 1

Write (k + 1)/2^k ≤ 1/8 or 8(k + 1)/2^k ≤ 1.

Then 8(k + 1) ≤ 2^k.

For 2^k − 8(k + 1) = 0, obtain d(2^k − 8k − 8)/dk equal to 2^k (ln 2) − 8.

By Newton's Method Of Root Approximation, establish the Formula

k − (2^k − 8k − 8)/[2^k (ln 2) − 8].

A graph of y = (k + 1)/2^k shows y equal to 1/8 around k = -0.9.

Create the Table below with the first value of k as -0.9; calculation 

of the Formula for each input of k is fed back into the formula as the 

new input value of k.

k-------------------------------k − (2^k − 8k − 8)/[2^k (ln 2) − 8]

(-0.9)------------------------------------------(-0.9346216794)

(-0.9346216794)---------------------------(-0.9346016356)

(-0.9346016356)---------------------------(-0.9346016356)

The last line of the Table shows the result of the Formula

equal to the input value of k for which the Formula was last

evaluated.

Inspection of a graph of y = (k + 1)/2^k will then show y less 

than or equal to 1/8 when k ≤ -0.9346016356.