Hints to compute if exists $\lim_{n\to\infty}\sum_{k=1}^n\sigma(k^2)/\sum_{k=1}^n\sigma(k)$, which $\sigma(n)=\sum_{d\mid n}d$, and other question I would like receive hints at least for one of the following problems, these are going from experiments. 

Can you provide to me hints for at least one of the following problems? I will try put the answer this week. If you want provide hints for one of the problems and solve the another, too is welcome. My goal is learn and made useful post for this site: 
A) Prove or refute that $\exists$ a real $\delta$ such that $$\lim_{n\to\infty}\sum_{k=1}^{n}\sigma(k)-\left(\frac{n}{rad(n)}\right)^{2+\delta}\sum_{k=1}^{n}\mu(k)k=0,$$
  where $\mu(m)$ is Mobius function, $\sigma(m)$ is the sum of divisors function and $rad(m)$ is defined by $rad(1)=1$ and if $m>1$ by the product of distinct primes dividing $m$, $\prod_{p\mid m}p$ (for example $rad(12)=6$).
I believe that previous exercise is more difficult than this
B) Compute, if exists, 
  $$\lim_{n\to\infty}\frac{\sum_{k=1}^n\sigma(k^2)}{\sum_{k=1}^n\sigma(k)}.$$

I believe that B) is more easy and useful currently to me. I know Bachmann's theorem about the average order of the sum of divisor function (page 60 in Apostol, Introduction to Analytic Number Theory), an how it was proved. If you want say an hint about the numerator in the limit of B), then I will try it. Thanks in advance.
I excuse that this post is tagged as experimental mathematics, since I've used my computer to claim this questions, I don't know if these are in the literature.
 A: 
Using the ideas in this answer Mean Value of a Multiplicative Function close to $n$ in Terms of the Zeta Function we can show that $$\sum_{n\leq x}\sigma(n)=\frac{\pi^2}{12}x^2+O(x)$$ $$\sum_{n\leq x} \sigma(n^2)=\frac{x^{3}}{6}\prod_{p}\left(1+\frac{p(p+1)}{p^{5}-1}\right)+O(x^{2})$$ 

Since $\sigma(n)=\sum_{d|n}d$, $\frac{\sigma(n)}{n}=\sum_{d|n}\frac{1}{d},$
 and so $$\sum_{n\leq x}\sigma(n)=\sum_{n\leq x}n\sum_{d|n}\frac{1}{d}=\sum_{d\leq x}\frac{1}{d}\sum_{\begin{array}{c}
n\leq x\\
d|n
\end{array}}n.$$
 Following the computations in Mean Value of a Multiplicative Function close to $n$ in Terms of the Zeta Function we then obtain $$\sum_{n\leq x}\sigma(n)=\frac{\pi^{2}}{12}x^{2}+O(x).$$
 To handle $\sum_{n\leq x}\sigma(n^2)$ lets try to write $$\frac{\sigma(k^{2})}{k^{2}}=(1*f)(k)$$
  for some $f$. Since $$\frac{\sigma\left(p^{2k}\right)}{p^{2k}}=1+p^{-1}+\cdots+p^{-2k},$$
 it follows that $$f(p^{k})=\begin{cases}
1 & k=1\\
p^{-2k+1}+p^{-2k} & k\geq1
\end{cases}.$$
 Thus $$\sum_{n\leq x}\sigma(n^{2})=\sum_{d\leq x}f(d)\sum_{\begin{array}{c}
n\leq x\\
d|n
\end{array}}n^{2}=\frac{1}{6}\sum_{d\leq x}f(d)\left[\frac{x}{d}\right]^{3}+O(x^{2}),$$
 and so $$\sum_{n\leq x}\sigma(n^{2}) = \frac{x^{3}}{6}\prod_{p}\left(1+\frac{f(p)}{p^{3}}+\frac{f(p^{2})}{p^{6}}+\cdots\right)+O(x^{2})
 = \frac{x^{3}}{6}\prod_{p}\left(1+\frac{p(p+1)}{p^{5}-1}\right)+O(x^{2}).$$
