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\section{Solving the sum}  \subsection{Laws and forumulas of summation}  These summation laws/formulas are used frequently.  They are very common and thus we won't prove any of them.  Most of them can be found \href{http://en.wikipedia.org/wiki/Summation}{here}  \begin{enumerate}  \item $\displaystyle\sum_{n=s}^t C\cdot f(n) = C\cdot \sum_{n=s}^t f(n)$  \item $\displaystyle\sum_{n=s}^t f(n) + \sum_{n=s}^{t} g(n) = \sum_{n=s}^t \left[f(n) + g(n)\right]$  \item $\displaystyle\sum_{n=s}^t f(n) - \sum_{n=s}^{t} g(n) = \sum_{n=s}^t \left[f(n) - g(n)\right]$  \item $\displaystyle\sum_{i=m}^n 1 = n - m + 1$  \item $\displaystyle\sum_{i=1}^n \frac{1}{i} = H_n$, where $H_n$ is the nth harmonic number.  \end{enumerate}  \subsection{Solving}  \begin{equation}  \begin{split}  m &= \sqrt{n}\\  T(n) &= a + \sum_{p = 2}^{m}\left[b + \sum_{i = 1}^{\frac{n}{p}} c\right]\\  &= a + \sum_{p = 2}^{m}b + \sum_{p = 2}^{m}\left[c\sum_{i = 1}^{\frac{n}{p}} 1\right]\\  &= a + b\sum_{p = 2}^{m}1 + c\sum_{p = 2}^{m}\left[\sum_{i = 1}^{\frac{n}{p}}\right]\\  &= a + b(m - 1) + c\sum_{p=2}^{m}\frac{n}{p}\\  &= a + b(m - 1) + cn\sum_{p=2}^{m}\frac{1}{p}\\  &= a + b(m - 1) + cn\left(\sum_{p=1}^{m}\frac{1}{p} - \sum_{p=1}^{1}\frac{1}{p}\right)\\  &= a + b(m - 1) + cn(H_m - 1)\\  &= a - b + bm + cnH_m - cn\\  &= a - b + b\sqrt{n} + cnH_\sqrt{n} - cn  \end{split}  \end{equation}  We know that $H_x$ oscillates around $ln(x)$, and that $H_x \in \mathcal{O}(log(n))$\\  Thus: \[T(n) \in \mathcal{O}(nlog(\sqrt{n}))\]  We use the logarithm identity $log_b(x^d) = dlog_b(x)$ to simplify and yield:  \[T(n) \in \mathcal{O}(nlog(n))\]