Simple equivalent of $\sum\limits_{i=1}^N\frac{\log i}i$ 
Find the limit $$\lim_{N \to \infty}\frac{\sum\limits_{i=1}^{N}\frac{\log i}{i}}{(\log N)^2}.$$

I am unable to find the sum $\sum\limits_{i=1}^{N}\frac{\log i}{i}$. Please help me.
EDIT: So,the limit is $$\lim_{N \to \infty}\frac{\frac{1}{2}\ln N^2}{(\ln N)^2}=\lim_{N \to \infty}\frac{\frac{1}{2}.2\ln N}{(\ln N)^2}=\lim_{N \to \infty}\frac{1}{\ln N}=0$$ I get the limit as $0$ not $\frac{1}{2}$. Please help. Thanks in advance.
 A: Notice that,
$$ \sum_{i=1}^{N}\frac{\ln(i)}{i}\sim\int_{1}^{N}\frac{\ln(x)}{x}dx = \frac{1}{2}\ln(N)^2=\frac{1}{2}(\ln(N))^2.$$
A: Let $a(n) = 1$ and $f(n) = \dfrac{\log(n)}{n}$, for all $n \in \mathbb{Z}^+$. We have that $A(t) = \displaystyle \sum_{n \leq t} a(n) = \lfloor t \rfloor$. Hence, $\displaystyle \sum_{n \leq x} \frac{\log n}{n} = \int_{1^-}^x f(t) dA(t) = \int_{1^-}^x \dfrac{\log(t)}{t} d (\lfloor t \rfloor)$. Integration by parts give us,
\begin{align*}
\int_{1^-}^x \frac{\log(t)}{t} d (\lfloor t \rfloor) & = \left( \frac{\log(t)}{t} \lfloor t \rfloor \right)_{1^-}^{x} - \int_{1^-}^{x} \frac{(1 - \log(t)) \lfloor t \rfloor}{t^2} dt\\
& = \log(x) - \int_{1^-}^{x} \frac{(1 - \log(t)) (t - \{t\})}{t^2} dt\\
& = \log(x) - \int_{1^-}^{x} \frac{(1 - \log(t))}{t} dt + \int_{1^-}^{x} \frac{(1 - \log(t)) \{t\}}{t^2} dt\\
& = \log(x) + \frac{\log^2(x)}{2} - \log(x) + \int_{1^-}^{x} \frac{(1 - \log(t)) \{t\}}{t^2} dt\\
& = \frac{\log^2(x)}{2} + \int_{1^-}^{\infty} \frac{(1 - \log(t)) \{t\}}{t^2} dt - \int_{x}^{\infty} \frac{(1 - \log(t)) \{t\}}{t^2} dt.
\end{align*}
Let $C = \displaystyle \int_{1^-}^{\infty} \frac{(1 - \log(t)) \{t\}}{t^2} dt$. Note that $\displaystyle \left \lvert C \right \rvert \leq \int_1^{\infty} \frac{1 + \log(t)}{t^2} dt = 2$.
Hence, we have that
\begin{align*}
\sum_{n \leq x} \frac{\log(n)}{n} = \frac{\log^2(x)}{2} + C + \int_{x}^{\infty} \frac{(\log(t)-1) \{t\}}{t^2} dt.
\end{align*}
Let $I = \displaystyle \int_x^{\infty} \frac{(\log(t) - 1) \{ t \}}{t^2} dt$. We will prove that $I = \mathcal{O} \displaystyle \left( \frac{\log(x)}{x} \right)$. Note that clearly $I > 0$ since $x$ is large and hence we can assume $x > e$. Further, we know that $0 \leq \{t\} < 1$. Hence, we have that $$I = \displaystyle \int_x^{\infty} \frac{(\log(t) - 1) \{ t \}}{t^2} dt < \int_x^{\infty} \frac{\log(t) -1}{t^2} dt = \int_x^{\infty} \frac{d}{dt} \left( -\frac{\log(t)}{t} \right) dt = \frac{\log(x)}{x}.$$
Hence, we get that
\begin{align*}
\sum_{n \leq x} \frac{\log(n)}{n} = \frac{\log^2(x)}{2} + C + \mathcal{O} \left( \frac{\log(x)}{x} \right).
\end{align*}
