There is this equation:
$$M = \sum\limits_{i = 1}^{\log n} {\frac{{in}}{{{2^i}}}} = n\sum\limits_{i = 1}^{\log n} {\frac{i}{{{2^i}}}} \leqslant n\sum\limits_{i = 1}^\infty {\frac{i}{{{2^i}}} = 2n} $$
And I don't understand why the rightmost summation can be simplified to 2n. Can you please explain it to me?