jmc

First, $$\sum _{k=0}^n{\log }_2\left(1+\frac{1}{2^{2^k}}\right)={\log }_2\left[\prod _{k=0}^n\left(1+\frac{1}{2^{2^k}}\right)\right].$$We want to evaluate $$(1+x)(1+x^2)(1+x^4)\cdots (1+x^{2^n})$$at $x=\frac{1}{2}.$ By difference of squares, $$\begin{array}{rl}(1+x)(1+x^2)(1+x^4)\cdots (1+x^{2^n})& =\frac{1-x^2}{1-x}\cdot \frac{1-x^4}{1-x^2}\cdot \frac{1-x^8}{1-x^4}\cdots \frac{1-x^{2^{n+1}}}{1-x^{2^n}}\\ & =\frac{1-x^{2^{n+1}}}{1-x}.\end{array}$$At $x=\frac{1}{2},$ $$\frac{1-x^{2^{n+1}}}{1-x}=\frac{1-(\frac{1}{2}{)}^{2^{n+1}}}{1-\frac{1}{2}}=2\left(1-\frac{1}{2^{2^{n+1}}}\right),$$and $${\log }_2\left[2\left(1-\frac{1}{2^{2^{n+1}}}\right)\right]={\log }_2\left(1-\frac{1}{2^{2^{n+1}}}\right)+1.$$Thus, we want the smallest positive integer $n$ such that $$1-\frac{1}{2^{2^{n+1}}}\ge \frac{2014}{2015}.$$This is equivalent to $$\frac{1}{2^{2^{n+1}}}\le \frac{1}{2015},$$or $2^{2^{n+1}}\ge 2015.$

For $n=2,$ $2^{2^{n+1}}=2^{2^3}=2^8=256,$ and for $n=3,$ $2^{2^{n+1}}=2^{2^4}=2^{16}=65536,$ so the smallest such $n$ is $\boxed{3}.$