Why do I love 2015 so much 1

$\begin{aligned} S & = \frac{\displaystyle \sum_{a=1}^{2015} \left(\displaystyle \sum_{b=1}^{2015} \frac{(-1)^{a-1}}{2b} \right)}{\displaystyle \sum_{c=1}^{2015}\frac{1}{c}} = \frac{\displaystyle \sum_{a=1}^{2015} \left(\frac{(-1)^{a-1}}{2} \displaystyle \sum_{b=1}^{2015} \frac{1}{b} \right)}{\displaystyle \sum_{c=1}^{2015}\frac{1}{c}} \\ & = \frac{\dfrac{1}{2} \displaystyle \sum_{a=1}^{2015} \left((-1)^{a-1} \displaystyle \sum_{b=1}^{2015} \frac{1}{b} \right)}{\displaystyle \sum_{c=1}^{2015}\frac{1}{c}} \quad \quad \small \color{#3D99F6}{\text{Let } S_k = \sum_{k=1}^{2015} \frac{1}{k}} \\ & = \frac{S_k-S_k+S_k-...-S_k+S_k}{2S_k} = \frac{S_k}{2S_k} = \frac{1}{2} \end{aligned}$

$\Rightarrow 4030S = \boxed{2015}$

I will provide you the very basic solution ,

$\frac{\displaystyle \sum^{2015}_{a=1} \sum^{2015}_{b=1} \frac{(-1)^{a-1}}{2b}} {\displaystyle \sum^{2015}_{c=1}\frac{1}{c}}$

Let's simplify the $Numerator$ of the above expression ,

$=\displaystyle \sum^{2015}_{a=1} \sum^{2015}_{b=1} \frac{(-1)^{a-1}}{2b}$

Now let's take a look of the $I^{st}$ term ,

$b=1 \Rightarrow$

$=\displaystyle \sum^{2015}_{a=1}\frac{(-1)^{a-1}}{2×1}$

$=\frac{1}{2}-\frac{1}{2}+\frac{1}{2}-...+\frac{1}{2}$

$=\frac{1}{2}$

$II^{nd}$ term ,

$b=2 \Rightarrow$

$=\displaystyle \sum^{2015}_{a=1}\frac{(-1)^{a-1}}{2×2}$

$=\frac{1}{2×2}-\frac{1}{2×2}+\frac{1}{2×2}-...+\frac{1}{2×2}$

$=\frac{1}{2×2}$

Therefore , $\displaystyle \sum^{2015}_{a=1} \sum^{2015}_{b=1} \frac{(-1)^{a-1}}{2b}$

$=\frac{1}{2}+\frac{1}{2×2}+\frac{1}{2×3}+...+\frac{1}{2×2015}$

$=\frac{1}{2}(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015})$

$=\frac{1}{2} \displaystyle \sum^{2015}_{k=1}\frac{1}{k}$

Now we can see that ,

$\displaystyle \sum^{2015}_{k=1}\frac{1}{k}=\displaystyle \sum^{2015}_{c=1}\frac{1}{c}$

Therefore,

$S=\frac{\displaystyle \sum^{2015}_{a=1} \sum^{2015}_{b=1} \frac{(-1)^{a-1}}{2b}} {\displaystyle \sum^{2015}_{c=1}\frac{1}{c}}=\frac{\frac{1}{2} \displaystyle \sum^{2015}_{k=1}\frac{1}{k}}{\displaystyle \sum^{2015}_{c=1}\frac{1}{c}}$

$S=\frac{1}{2}$

Therefore , $4030S=\frac{4030}{2}$

$=\boxed{2015.000}$

$S=\dfrac{\displaystyle\sum_{a=1}^{2015}(\displaystyle\sum_{b=1}^{2015}\frac{(-1)^{a-1}}{2b})}{\displaystyle\sum_{c=1}^{2015}\frac{1}{c}}$ $=\frac{1}{2}\dfrac{(\displaystyle\sum_{b=1}^{2015}\frac{1}{b})(\displaystyle\sum_{a=1}^{2015}(-1)^{a-1})}{\displaystyle\sum_{c=1}^{2015}\frac{1}{c}}$ $\frac{1}{2}\cdot (1)=\frac{1}{2}$