Computing the integral $\int_{1}^{2} \left( \sum_{k=0}^{\infty} \frac{k}{(2x)^{k+1}} \right)dx$ I'm having a hard time computing the following integral
$$\int_{1}^{2} \left( \sum_{k=0}^{\infty} \frac{k}{(2x)^{k+1}} \right)dx$$. 
Well not hard time computing its value (haha, WolframAlpha), but I have a hard time doing it on paper by hand. I usually do not understand which methods and ways can I use and most importantly why can I do that... 
This springs' exam period has hit me a bit harder than usual so I hope someone can help me even though the problem itself is probably very easy and my question about the used methods and their reasons just foolish...
 A: Since switching the sum and the integral signs would be the fastest way to proceed, we'll start by proving that we can do so :
\begin{aligned}\left|\sum_{k=0}^{n}{\int_{1}^{2}{\frac{k}{\left(2x\right)^{k+1}}\,\mathrm{d}x}}-\int_{1}^{2}{\sum_{k=0}^{+\infty}{\frac{k}{\left(2x\right)^{k+1}}}\,\mathrm{d}x}\right|&=\int_{1}^{2}{\sum_{k=n+1}^{+\infty}{\frac{k}{\left(2x\right)^{k+1}}}\,\mathrm{d}x}\\ &\leq\sum_{k=n+1}^{+\infty}{\frac{k}{2^{k+1}}}\underset{n\to +\infty}{\longrightarrow}0\end{aligned}
That's because $ \sum\limits_{n\geq 0}{\frac{n}{2^{n+1}}} $ converges so the sequence of remains $ \left(\sum\limits_{k=n+1}^{+\infty}{\frac{k}{2^{k+1}}}\right)_{n} $ will have $ 0 $ as a limit.
Thus : $$ \sum_{k=0}^{+\infty}{\int_{1}^{2}{\frac{k}{\left(2x\right)^{k+1}}}\,\mathrm{d}x}=\lim_{n\to +\infty}{\sum_{k=0}^{n}{\int_{1}^{2}{\frac{k}{\left(2x\right)^{k+1}}\,\mathrm{d}x}}}=\int_{1}^{2}{\sum_{k=0}^{+\infty}{\frac{k}{\left(2x\right)^{k+1}}}\,\mathrm{d}x} $$
Hence : \begin{aligned}\int_{1}^{2}{\sum_{k=0}^{+\infty}{\frac{k}{\left(2x\right)^{k+1}}}\,\mathrm{d}x}&=\sum_{k=0}^{+\infty}{\int_{1}^{2}{\frac{k}{\left(2x\right)^{k+1}}}\,\mathrm{d}x}\\ &=\sum_{k=1}^{+\infty}{\frac{k}{2^{k+1}}\int_{1}^{2}{\frac{\mathrm{d}x}{x^{k+1}}}}\\ &=\sum_{k=1}^{+\infty}{\frac{1-2^{-k}}{2^{k+1}}}\\ &=\frac{1}{2}\left(\sum_{k=1}^{+\infty}{\frac{1}{2^{k}}}-\sum_{k=1}^{+\infty}{\frac{1}{4^{k}}}\right)\\&=\frac{1}{3}\end{aligned}
A: Integral doesn't depend on $k$, so you can rewrite the expression as 
$$
\sum_{k=0}^{\infty} k\int_{1}^{2}(2x)^{-(k+1)}dx
$$
Can you handle from here?
