Chapter 4, Problem 39C

Explanation of Solution

Given:

It is given that ${\displaystyle \sum _{i=1}^n\frac{i}{2^i}}<2$

Proof:

Let us prove the above case by induction steps.

Base case: Let us assume that the “n” value is “0”. Substitute the value in the given integral results in:

$S={\displaystyle \sum _{i=1}^0\frac{1}{2^1}}<2$

Thus, from the above integral the value of “S” is less than 2.

Induction case:

In induction case, take the following summation:

$S=\left({\displaystyle \sum _{i=1}^n\frac{1}{2^i}}\right)+({\displaystyle \sum _{i=2}^n\frac{\left(i-1\right)}{2^i}}$