5. For each positive integer $n$, let $\varphi(n)$ be the number of positive integers from 1 to $n$ that are relatively prime to $n$. Evaluate
$$
\sum_{n=1}^{\infty} \frac{\varphi(n) 4^{n}}{7^{n}-4^{n}}
$$
Answer: $\frac{28}{9}$
Solution: We compute
$$
\sum_{n=1}^{\infty} \frac{\varphi(n) 4^{n}}{7^{n}-4^{n}}=\sum_{n=1}^{\infty} \varphi(n) \frac{\left(\frac{4}{7}\right)^{n}}{1-\left(\frac{4}{7}\right)^{n}}=\sum_{n=1}^{\infty} \varphi(n) \sum_{k=1}^{\infty}\left(\frac{4}{7}\right)^{n k}
$$

Interchanging the order of summation and using the fact that $\sum_{d \mid n} \varphi(d)=n$ where the sum takes all over positive divisors of $n$, we arrive at
$$
\begin{aligned}
\sum_{n=1}^{\infty} \frac{\varphi(n) 4^{n}}{7^{n}-4^{n}} & =\sum_{n=1}^{\infty} \varphi(n) \sum_{k=1}^{\infty}\left(\frac{4}{7}\right)^{n k}=\sum_{m=1}^{\infty}\left(\frac{4}{7}\right)^{m} \sum_{d \mid m} \varphi(d) \\
& =\sum_{m=1}^{\infty} m\left(\frac{4}{7}\right)^{m}=\frac{\frac{4}{7}}{\left(1-\frac{4}{7}\right)^{2}}=\frac{28}{9}
\end{aligned}
$$