Harmonic sum inequality

The Key to solve this question is to realise that this inequality holds for each and every Natural Number.

If we closely examine the harmonic series and round the denominator of the fractions to the closest larger and smaller power of 2 [Example $\frac{1}{3}$ becomes $\frac{1}{4}$ when rounding to the larger power and $\frac{1}{2}$ when rounding to the smaller power] we obtain $2$ series.The sum of the first series will always be smaller then the harmonic series and the second series will always be greater except at $n = 1$ where all of them are equal.Thus we obtain the inequality -

$\frac{1}{1} + \frac{1}{2} + \frac{1}{4} + \frac{1}{4} ................. \leq \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} .....................\leq \frac{1}{1} + \frac{1}{2} + \frac{1}{2} + \frac{1}{4}....................$

Note that the sum of the first series can be represented by the expression $\log_{2}{2\sqrt{n}}$ . Similarly the sum of the second series can be represented by the the expression $\log_{2}{n + 1}$ .Thus the above inequality transforms into -

$\log_{2}{2\sqrt{n}} \leq\sum_1^n \frac{1}{i} \leq\log_{2}{n+1}$

Thus the above inequality is satisfies by all the natural numbers.Thus the sum of the first,third and fifth solution is -

$1 + 3 + 5 = 9$