Define $a_n=\frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{2n}$ and note that $a_n-a_{n-1}=\frac{1}{2n(2n-1)}$, so $a_n=\sum_{k=1}^n\frac{1}{2k(2k-1)}<\sum_{k=1}^\infty\frac{1}{2k(2k-1)}=\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k}=\ln 2<0.7$. Also, $a_{1008}=\sum_{k=1}^{1008}\frac{1}{2k(2k-1)}>\frac{1}{2}+\frac{1}{12}+\frac{1}{30}>0.6$, so the answer is 6.

Let $S=\sum_{i=1009}^{2016}{\frac{1}{i}}$, we want to compute $T=\lfloor 10S\rfloor$ Note that $S=\sum_{i=1}^{2016}{\frac{1}{i}}-\sum_{i=1}^{1008}{\frac{1}{i}}=\sum_{i=1}^{2016}{\frac{1}{i}}-2\sum_{i=1}^{1008}{\frac{1}{2i}}=\sum_{i=1}^{1008}{\Big( \frac{1}{2i-1}-\frac{1}{2i}\Big)} =\sum_{i=1}^{1008}{\frac{1}{(2i-1)2i}}$ So $10S=5+\frac{7}{6}+10\Big(\frac{1}{7\times 8}+\frac{1}{9\times 10}+...+\frac{1}{2015\times 2016}\Big)<6+\frac{1}{6}+10\left( \frac{1}{56}+\frac{1}{90}+\frac{1}{132}\right)+10\times \Big( \sum_{i=6}^{\infty}{\frac{1}{(2i)^2}}\Big) =6+\frac{1}{6}+2.5\left( \frac{\pi^2}{6}-1-\frac{1}{4}-\frac{1}{9}-\frac{1}{16}-\frac{1}{25}\right)+10\left( \frac{1}{56}+\frac{1}{90}+\frac{1}{132}\right)<7$ So $6<10S<7$, we get $T=6$

Very easy one! By C-S we have that $\frac{1}{1009+i}+\frac{1}{2016-i}>\frac{4}{3017}$, so that means our sum $S=\sum_{i=1009}^{2016}{\frac{1}{i}}>\frac{4\times504}{3017}>0.6$. On the other side we have that $1009\times2016<(1009+i)(2016-i)$, where $0<i<505$. Now grouping all the factors by $2$ such that the denominators in sum give $3017$ we get that sum $S=\sum_{i=1009}^{2016}{\frac{1}{i}}<\frac{504\times3017}{1009\times2016}<0.7$. This means the answer is $6$.

Probably one of the first problems I've solved with Calculus. And I don't really know Calculus... From what it seems looks like, I feel like this should almost be a routine approach? Note \begin{align*} \frac{1}{1009}+\frac{1}{1010}+\dots+\frac{1}{2016} &= \left( \frac{1}{1009}+\frac{1}{2016} \right) + \left( \frac{1}{1010}+\frac{1}{2015} \right)+\dots+\left( \frac{1}{1512}+\frac{1}{1513} \right) \\ & \geq \frac{1006}{2} \cdot \frac{4}{3025} \\ & > 0.6 \end{align*}by AM-HM, as $\tfrac{1}{a}+\tfrac{1}{b} \geq \tfrac{4}{a+b}$. On the other hand by Riemann Integration \begin{align*} \frac{1}{1009}+\frac{1}{1010}+\dots+\frac{1}{2016} &< \int_{1008}^{2016} \frac{1}{x} \, \mathrm{d}x \\ &= \ln{2016} - \ln{1008} \\ &= \ln{2}. \end{align*}It suffices to show $\ln 2 < 0.7$. That is equivalent to showing $2 < e^{0.7}$. But $e^{0.7} > 2.71^{0.7} >2$ is easy to show by computing. Therefore, we are done and the first decimal digit of the discussed sum is $6$.

Euler was the one that found that the value of $\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + ... + \frac{1}{n} = k + \ln{n}$. Where $k$ is a constant with approximated value of $0.58$. (Oh, if this is a theorem and has a specific name, I would love to know). Now we have that that sum is: $\frac{1}{1} + ... + \frac{1}{2016} - (\frac{1}{1} + ... + \frac{1}{1008}) = k + \ln{2016} - k - \ln{1008} = \ln{2}$ We only have to know that $ln{2} \approx 0.6...$ Thus the answer is $\boxed{6}$.

@above: Actually \[ H_n := \sum_{i=1}^{n} \frac{1}{i} \approx \gamma+\ln{n} \]instead of your equal sign. The constant $\gamma$ indeed has a name, it is the so called Euler–Mascheroni constant. It is evident that the equal sign cannot be correct as that'd contradict the proof of myself or jeff10.

Oh I get it. Thanks for the constant name and the correction!