Let $f(x)$ be the greatest prime number at most $x$. Let $g(x)$ be the least prime number greater than $x$. Find $$\sum_{i=2}^{100}\frac{1}{f(i)g(i)}.$$
We could just computationally calculate the sum $$\sum_{i=2}^{100}\frac{1}{f(i)g(i)} = \dfrac{1}{2 \cdot 3}+\dfrac{1}{3 \cdot 5}+\dfrac{1}{3 \cdot 5}+\cdots+\dfrac{1}{97 \cdot 101},$$ but is there a way to avoid all the computations?
Denote $p_n$ be the $n^{th}$ prime. Then there always exist $k$ such that
$$p_{k-1}=f(i)\le i Notice
$$\frac1{f(i)g(i)}=\frac1{p_{k-1}p_{k}}=
\frac1{p_{k}-p_{k-1}}\cdot(\frac1{p_{k-1}}-\frac1{p_{k}})$$ By observation we know for every $i$, the number of $\displaystyle\frac1{p_{k-1}p_{k}}$ terms is $p_{k}-p_{k-1}$. Thus the sum of these terms is $\displaystyle\frac1{p_{k-1}}-\frac1{p_{k}}$. Therefore \begin{align}
\sum_{i=2}^{100}\frac{1}{f(i)g(i)}&=\sum_{p_k\le 97}(\frac1{p_{k-1}}-\frac1{p_{k}})+\sum_{i=97}^{100}\frac{1}{f(i)g(i)}\\
&=(\frac12-\frac1{97})+4\cdot\frac1{97\cdot101}\\
&=\frac{99}{202}
\end{align}