Prove a sum of fractions less than a value

Question

I happened to see this problem from an elementary school textbook, but cannot solve it:

$$ C = \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + ... + \frac{1}{15} + \frac{1}{16} + \frac{1}{17} $$

Prove $$C < 2$$

Very much appreciate someone to enlighten me, especially with an elementary solution.

Answer 1

$$ C = \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + ... + \frac{1}{15} + \frac{1}{16} + \frac{1}{17} $$ $$C=\frac15+\frac16+\frac17+\frac18+\frac19+\frac1{10}+\frac1{11}+\frac1{12}+\frac1{13}+\frac1{14}+\frac1{15}+\frac1{16}+\frac1{17}\le\left(\frac15\times6\right)+\left(\frac1{10}\times7\right)=\frac{19}{10}\lt2$$

Answer 2

they are usually comparing denominators to powers of two;

sum(5,6,7) $ < 3/4$

sum(8,9,10,11,12,13,14,15) $< 8/8 = 1$

sum(16,17) $< 2/16 = 1/8$

$$ C < \frac{15}{8} < 2 $$

In the other direction:

sum(5,6,7,8) $ > 4/8 = 1/2$

sum(9,10,11,12,13,14,15,16) $> 8/16 = 1/2$

$$ C > 1$$

Answer 3

It is easily seen that the terms $\frac{1}{5},\frac{1}{6},\frac{1}{7},\frac{1}{8},\frac{1}{9}$ all are less than or equal to $\frac{1}{5},$ so the sum of them is less than $1$ and similarly the sum of other is less than $1$ also all of them is less than or equal to $\frac{1}{10}$ then $C<2.$

Answer 4

$$\begin{align} C &\le 3\times\frac15 + 10\times\frac18\\ \implies C &\le \frac35 + \frac54\\ \implies C &\le \frac {12}{20} + \frac{25}{20}\\ \implies C &\le \frac{37}{20}\le \frac {40}{20} = 2 \end{align}$$