15. $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots+\frac{1}{2 n-1}-\frac{1}{2 n}=\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2 n}$. By the Cauchy-Schwarz inequality, we have $[(n+1)+(n+2)+\cdots+(2 n)]\left(\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2 n}\right)>$ $n^{2}$, so $\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2 n}>\frac{2 n}{3 n+1} \geqslant \frac{4}{7}$. Also, $\frac{1}{n+1}+$ $\frac{1}{n+2}+\cdots+\frac{1}{2 n}<\left(1^{2}+1^{2}+\cdots+1^{2}\right)^{\frac{1}{2}}\left[\frac{1}{(n+1)^{2}}+\cdots+\frac{1}{(2 n)^{2}}\right]^{\frac{1}{2}}<$ $\sqrt{n}\left[\frac{1}{n(n+1)}+\frac{1}{(n+1)(n+2)}+\cdots+\frac{1}{(2 n-1) \cdot 2 n}\right]^{\frac{1}{2}}=\frac{\sqrt{2}}{2}$