Is $\frac14 + \frac15 + \cdots + \frac1{14} + \frac1{15}\lt\frac32$? Use some combination of arithmetic, algebra, and/or elementary integral calculus to determine if the sum
$$\frac14 + \frac15 + \cdots + \frac1{14} + \frac1{15}\lt\frac32$$
 A: We can make use of the serendipitous fact that the product of the three cumbersome primes $7$, $11$ and $13$ is almost exactly $1000$:
$$
\frac17+\frac1{11}+\frac1{13}=\frac{11\cdot13+13\cdot7+7\cdot11}{1001}=\frac{311}{1001}\lt\frac{311}{1000}\;.
$$
The remaining fractions can be combined like this:
$$
\frac14+\frac18=\frac{3}{8}=\frac{375}{1000}\;,
$$
$$
\frac15+\frac1{10}=\frac3{10}=\frac{300}{1000}\;,
$$
\begin{align}
\frac16+\frac19+\frac1{12}+\frac1{14}+\frac1{15}&\lt\frac16+\frac19+\frac1{12}+\frac1{12}+\frac1{15}\\
&=\frac{30+20+15+15+12}{180}\\
&=\frac{92}{180}\\
&=\frac12+\frac1{90}
\end{align}
The first three sums sum to
$$
\frac{311+375+300}{1000}=\frac{986}{1000}=1-\frac{14}{1000}\;.
$$
Then the desired result follows from
$$
\frac1{90}-\frac{14}{1000}=\frac{100-9\cdot14}{9000}=-\frac{26}{9000}\;.
$$
I'm afraid I haven't used any algebra or calculus, but it does say "and/or". :-)
A: As $\frac1x$ is convex, we get
$$\frac1k \le \int_{k-\frac12}^{k+\frac12}\frac1x dx \quad \implies \sum_{k=4}^{15}\frac1k \le \int_{3.5}^{15.5}\frac1xdx = \log\frac{31}7 < \frac32$$
Of course that still needs you to approximate $\log \frac{13}7$, though on the positive side it would work well for much larger number of terms...
