a) Find the value of the sum

$$
\frac{1}{1+1 / x}+\frac{1}{1+x}
$$

b) Find the value of the sum

$$
\frac{1}{2019^{-2019}+1}+\ldots+\frac{1}{2019^{-1}+1}+\frac{1}{2019^{0}+1}+\frac{1}{2019^{1}+1}+\ldots+\frac{1}{2019^{2019}+1}
$$
Solution

a) We have

$$
\begin{aligned}
\frac{1}{1+1 / x}+\frac{1}{1+x} & =\frac{1}{(x+1) / x}+\frac{1}{1+x} \\
& =\frac{x}{1+x}+\frac{1}{1+x} \\
& =1
\end{aligned}
$$

b) Due to the previous item, considering $x=a^{b}$, we can group the fractions $\frac{1}{a^{-b}+1}$ and $\frac{1}{a^{b}+1}$ in pairs that sum to 1. Removing the term $\frac{1}{2019^{0}+1}=1 / 2$, we can rewrite the sum of the remaining terms as

$$
\begin{aligned}
& \left(\frac{1}{2019^{-2019}+1}+\frac{1}{2019^{2019}+1}\right)+\left(\frac{1}{2019^{-2018}+1}+\frac{1}{2019^{2018}+1}\right)+ \\
& \left(\frac{1}{2019^{-2017}+1}+\frac{1}{2019^{2017}+1}\right)+\left(\frac{1}{2019^{-2016}+1}+\frac{1}{2019^{2016}+1}\right)+\ldots
\end{aligned}
$$

The sum of these 2019 pairs is 2019. Thus, the requested sum is $2019+1 / 2=\frac{4039}{2}$.