The answer to Question 1 is "yes". To see this, notice that $s_n$ is at most square root of the average square of divisor, i.e. $$ s_n\leq \sqrt{\frac{\sum_{d\mid n}d^2}{\sum_{d\mid n} 1}}=\sqrt{\frac{\sigma_2(n)}{\sigma_0(n)}}, $$

where $\sigma_k(n)$ is the sum of $k$-th powers of divisors of $n$. Now,

$$ \sigma_2(n)=n^2\sigma_{-2}(n), $$

so

$$ \sigma_2(n)<\frac{\pi^2}{6}n^2 $$

for all $n$. Therefore we have

$$ f(n)\leq \frac{2}{n-1} \sqrt{\frac{\pi^2}{6}n^2/\sigma_0(n)}\leq \frac{5.14}{\sqrt{\sigma_0(n)}} $$

for all $n$. Now, almost all $n\leq N$ have at least $0.5\ln\ln N$ distinct prime factors. In particular, for almost all $n\leq N$ we have $\sigma_0(n)\geq 0.5\ln\ln N$. Therefore, our bound for $f(n)$ together with the trivial observation that $0\leq f(n)\leq 1$ gives

$$ \sum_{n\leq N} f(n)\leq \sum_{n\leq N, \sigma_0(n)\geq 0.5\ln\ln N} \frac{5.14}{\sigma_0(n)}+\sum_{n\leq N, \sigma_0(n)<0.5\ln\ln N} 1= o(N), $$

as needed.

Using contour integration method one can even prove something like

$$ \sum_{n\leq N} f(n)=O(N(\ln N)^{1/\sqrt{2}-1}) $$

The answer to Question 1 is "yes". To see this, notice that $s_n$ is at most square root of the average square of divisor, i.e. $$ s_n\leq \sqrt{\frac{\sum_{d\mid n}d^2}{\sum_{d\mid n} 1}}=\sqrt{\frac{\sigma_2(n)}{\sigma_0(n)}}, $$

where $\sigma_k(n)$ is the sum of $k$-th powers of divisors of $n$. Now,

$$ \sigma_2(n)=n^2\sigma_{-2}(n), $$

so

$$ \sigma_2(n)<\frac{\pi^2}{6}n^2 $$

for all $n$. Therefore we have

$$ f(n)\leq \frac{2}{n-1} \sqrt{\frac{\pi^2}{6}n^2/\sigma_0(n)}\leq \frac{5.14}{\sqrt{\sigma_0(n)}} $$

for all $n$. Now, almost all $n\leq N$ have at least $0.5\ln\ln N$ distinct prime factors. In particular, for almost all $n\leq N$ we have $\sigma_0(n)\geq 0.5\ln\ln N$. Therefore, our bound for $f(n)$ together with the trivial observation that $0\leq f(n)\leq 1$ gives

$$ \sum_{n\leq N} f(n)\leq \sum_{n\leq N, \sigma_0(n)\geq 0.5\ln\ln N} \frac{5.14}{\sqrt{\sigma_0(n)}}+\sum_{n\leq N, \sigma_0(n)<0.5\ln\ln N} 1= o(N), $$

as needed.

Using contour integration method one can even prove something like

$$ \sum_{n\leq N} f(n)=O(N(\ln N)^{1/\sqrt{2}-1}) $$

The answer to Question 1 is "yes". To see this, notice that $s_n$ is at most square root of the average square of divisor, i.e. $$ s_n\leq \sqrt{\frac{\sum_{d\mid n}d^2}{\sum_{d\mid n} 1}}=\sqrt{\frac{\sigma_2(n)}{\sigma_0(n)}}, $$

where $\sigma_k(n)$ is the sum of $k$-th powers of divisors of $n$. Now,

$$ \sigma_2(n)=n^2\sigma_{-2}(n), $$

so

$$ \sigma_2(n)<\frac{\pi^2}{6}n^2 $$

for all $n$. Therefore we have

$$ f(n)\leq \frac{2}{n-1} \sqrt{\frac{\pi^2}{6}n^2/\sigma_0(n)}\leq \frac{5.14}{\sqrt{\sigma_0(n)}} $$

for all $n$. Now, almost all $n\leq N$ have at least $0.5\ln\ln N$ distinct prime factors. In particular, for almost all $n\leq N$ we have $\sigma_0(n)\geq 0.5\ln\ln N$. Therefore, our bound for $f(n)$ together with the trivial observation that $0\leq f(n)\leq 1$ gives

$$ \sum_{n\leq N} f(n)\leq \sum_{n\leq N, \sigma_0(n)\geq 0.5\ln\ln N} \frac{5.14}{\sigma_0(n)}+\sum_{n\leq N, \sigma_0(n)<0.5\ln\ln N} 1= o(N), $$

as needed.

Using contour integration method one can even prove something like

$$ \sum_{n\leq N} f(n)=O\left(\frac{N}{(\ln N)^{1/\sqrt{2}}}\right) $$

The answer to Question 1 is "yes". To see this, notice that $s_n$ is at most square root of the average square of divisor, i.e. $$ s_n\leq \sqrt{\frac{\sum_{d\mid n}d^2}{\sum_{d\mid n} 1}}=\sqrt{\frac{\sigma_2(n)}{\sigma_0(n)}}, $$

where $\sigma_k(n)$ is the sum of $k$-th powers of divisors of $n$. Now,

$$ \sigma_2(n)=n^2\sigma_{-2}(n), $$

so

$$ \sigma_2(n)<\frac{\pi^2}{6}n^2 $$

for all $n$. Therefore we have

$$ f(n)\leq \frac{2}{n-1} \sqrt{\frac{\pi^2}{6}n^2/\sigma_0(n)}\leq \frac{5.14}{\sqrt{\sigma_0(n)}} $$

for all $n$. Now, almost all $n\leq N$ have at least $0.5\ln\ln N$ distinct prime factors. In particular, for almost all $n\leq N$ we have $\sigma_0(n)\geq 0.5\ln\ln N$. Therefore, our bound for $f(n)$ together with the trivial observation that $0\leq f(n)\leq 1$ gives

$$ \sum_{n\leq N} f(n)\leq \sum_{n\leq N, \sigma_0(n)\geq 0.5\ln\ln N} \frac{5.14}{\sigma_0(n)}+\sum_{n\leq N, \sigma_0(n)<0.5\ln\ln N} 1= o(N), $$

as needed.

Using contour integration method one can even prove something like

$$ \sum_{n\leq N} f(n)=O(N(\ln N)^{1/\sqrt{2}-1}) $$