If $n = p_1^{\alpha_1} \cdot \ldots \cdot p_k^{\alpha_k}$, where $p_1$ is lowest prime divisor.
Let $d(n) = \alpha_1$.
Is there any upper bound for $\displaystyle \sum_i^{n} d(i)$?
We can roughly estimate it as $O(n \log n)$, because $d(n) \in O(\log n)$, but maybe exists better estimate for this sum? For example $O(n \log \log n)$?
Best Answer
We get an $O(n \log \log n)$ bound from the obvious estimate $$d(n) = \alpha_1 \leqslant \alpha_1 + \ldots + \alpha_k = \Omega(n)$$ and the rather well-known (e.g. Theorem 430 in Hardy/Wright) $$\sum_{n \leqslant x} \Omega(n) = x\log \log x + B_2x + o(x)$$ which follows from Mertens' second theorem.
We can get an asymptotically smaller bound, $O(n)$, by using a slightly more conservative bound. With $\omega(n)$ denoting the number of distinct prime factors of $n$, i.e. $\omega(n) = k$ if $n = p_1^{\alpha_1}\ldots p_k^{\alpha_k}$, we can easily see $$d(n) = \alpha_1 \leqslant 1 + (\alpha_1 - 1) + \ldots + (\alpha_k - 1) = 1 + \Omega(n) - \omega(n)\,.$$ Thus, since (see also Theorem 430 in Hardy/Wright) $$\sum_{n \leqslant x} \omega(n) = x\log \log x + B_1 x + o(x)$$ we have \begin{align} \sum_{i = 1}^{n} d(i) &\leqslant \sum_{i = 1}^{n} \bigl(1 + \Omega(i) - \omega(i)\bigr) \\ &= n + \sum_{i = 1}^{n} \Omega(i) - \sum_{i = 1}^{n}\omega(i) \\ &= n + (B_2 - B_1)n + o(n)\,. \end{align} Since trivially $d(i) \geqslant 1$ for $i > 1$ it follows that $$\sum_{i = 1}^{n} d(i) \asymp n\,.$$