4. (5 points) If $\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{2^{n}}>\frac{315}{412}$ ( $n$ is a natural number greater than 0), then the smallest value of $n$ that satisfies the condition is . $\qquad$
4. (5 points) If $\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{2^{\mathrm{n}}}>\frac{315}{412}$ ( $n$ is a natural number greater than 0), then the smallest value of $n$ that satisfies the condition is $\qquad$
【Solution】Solution: When $n=1$, the left side of the inequality equals $\frac{1}{2}$, which is less than $\frac{315}{412}$, and does not satisfy the condition;
When $n=2$, the left side of the inequality equals $\frac{1}{2}+\frac{1}{4}=\frac{3}{4}=\frac{309}{412}$, which is less than $\frac{315}{412}$, and does not satisfy the condition;
Similarly, when $n=3$, the left side of the inequality is greater than $\frac{315}{412}$, and satisfies the condition;
Therefore, the smallest value of $n$ that satisfies the condition is 3.
So the answer is: 3