3. Given an arithmetic sequence $\left\{a_{n}\right\}$ with a common difference of $d$ satisfying $d>0$, and $a_{2}$ is the geometric mean of $a_{1}$ and $a_{4}$. Let $b_{n}=a_{2^{*}}\left(n \in \mathbf{Z}_{+}\right)$, for any positive integer $n$ we have
$$
\frac{1}{b_{1}}+\frac{1}{b_{2}}+\cdots+\frac{1}{b_{n}}<2 \text {. }
$$

Then the range of the common difference $d$ is ( ).
(A) $\left[\frac{1}{2},+\infty\right)$
(B) $\left(\frac{1}{2},+\infty\right)$
(C) $\left[\frac{1}{4}, \frac{1}{2}\right)$
(D) $\left[\frac{1}{4}, \frac{1}{2}\right]$
3. A.

From the problem, we know
$$
\begin{array}{l}
a_{2}^{2}=a_{1} a_{4} \Rightarrow\left(a_{1}+d\right)^{2}=a_{1}\left(a_{1}+3 d\right) \\
\Rightarrow a_{1}=d .
\end{array}
$$

Thus, $a_{n}=n d$. Therefore, $b_{n}=a_{2^{*}}=2^{n} d$. Then $\frac{1}{d}\left[\frac{1}{2}+\left(\frac{1}{2}\right)^{2}+\cdots+\left(\frac{1}{2}\right)^{n}\right]=\frac{1}{2}\left[1-\left(\frac{1}{2}\right)^{n}\right]$ holds for any positive integer $n$. Therefore, $d \geqslant \frac{1}{2}$.