题目（高二 · 算术）：23. (2015 广东) 数列 $\left\{a_{n}\right\}$ 满足: $a_{1}+2 a_{2}+\cdots+n a_{n}=4-\frac{n+2}{2^{n-1}}, n \in \mathrm{N}^{*}$.

(1) 求 $a_{3}$ 的值;

(2) 求数列 $\left\{a_{n}\right\}$ 的前 $n$ 项和 $T_{n}$;

（3）令 $b_{1}=a_{1}, b_{n}=\frac{T_{n-1}}{n}+\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}\right) a_{n}(n \geqslant 2)$

证明: 数列 $\left\{b_{n}\right\}$ 的前 $n$ 项和 $S_{n}$ 满足 $S_{n}<2+2 \ln n$.
选项：
答案：（1）由题意知: $a_{1}+2 a_{2}+\cdots+n a_{n}=4-\frac{n+2}{2^{n-1}}$

当 $n=3$ 时, $a_{1}+2 a_{2}=4-\frac{2+2}{2^{1}}$;

当 $n=3$ 时, $a_{1}+2 a_{2}+3 a_{3}=4-\frac{3+2}{2^{2}}$;

$3 a_{3}=4-\frac{3+2}{2^{2}}-\left(4-\frac{2+2}{2^{1}}\right)=\frac{3}{4}$

$a_{3}=\frac{1}{4}$

(2) 当 $n=1$ 时, $a_{1}=4-\frac{1+2}{2^{1-1}}=1$;

当 $n \geqslant 2$ 时, 由 $a_{1}+2 a_{2}+\cdots+n a_{n}=4-\frac{n+2}{2^{n-1}}$ 知

$a_{1}+2 a_{2}+\cdots+(n-1) a_{n-1}=4-\frac{n-1+2}{2^{n-2}}$

两式相减得 $n a_{n}=\frac{n+1}{2^{n-2}}-\frac{n+2}{2^{n-1}}=\frac{n}{2^{n-1}}$, 此时 $a_{n}=\frac{1}{2^{n-1}}$.

经检验知 $a_{1}=1$ 也满足 $a_{n}=\frac{1}{2^{n-1}}$. 故数列 $\left\{a_{n}\right\}$ 是以 1 为首项, $\frac{1}{2}$ 为公比的公比数列,

故 $T_{n}=\frac{1 \times\left[1-\left(\frac{1}{2}\right)^{n}\right]}{1-\frac{1}{2}}=2-\frac{1}{2^{n-1}}$.

(3) 由（1）(2) 知, $b_{1}=a_{1}=1$.

当 $n \geqslant 2$ 时, $b_{n}=\frac{T_{n-1}}{n}+\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}\right) a_{n}=\frac{2-\frac{1}{2^{n-2}}}{n}+\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}\right) \cdot \frac{1}{2^{n-1}}$ $=\frac{2}{n}+\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n-1}-\frac{1}{n}\right) \cdot \frac{1}{2^{n-1}}$.

当 $n=1$ 时, $S_{1}=1<2+2 \ln 1=2$, 成立;

当 $n \geqslant 2$ 时,

$$
\begin{aligned}
S_{n}= & +\left[\frac{2}{2}+\left(1-\frac{1}{2}\right) \cdot \frac{1}{2^{1}}\right]+\left[\frac{2}{3}+\left(1+\frac{1}{2}-\frac{1}{3}\right) \cdot \frac{1}{2^{2}}\right]+\cdots \\
& +\left[\frac{2}{n}+\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n-1}-\frac{1}{n}\right) \cdot \frac{1}{2^{n-1}}\right] \\
= & 1+2\left(\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}\right)+\left(\frac{1}{2}+\frac{1}{2^{2}}+\cdots+\frac{1}{2^{n-1}}\right)+\frac{1}{2}\left(\frac{1}{2^{2}}+\frac{1}{2^{3}}+\cdots+\frac{1}{2^{n-1}}-\frac{1}{2}\right) \\
& +\frac{1}{3}\left(\frac{1}{2^{3}}+\frac{1}{2^{4}}+\cdots+\frac{1}{2^{n-1}}-\frac{1}{2^{2}}\right)+\cdots+\frac{1}{n-1}\left(\frac{1}{2^{n-1}}-\frac{1}{2^{n-2}}\right)+\frac{1}{n}\left(-\frac{1}{2^{n-1}}\right) \\
= & 1+2\left(\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}\right)+\left(1-\frac{1}{2^{n-1}}\right)+\frac{1}{2}\left(\frac{1}{2^{2}} \cdot \frac{1-\frac{1}{2^{n-2}}}{1-\frac{1}{2}}-\frac{1}{2}\right) \\
& +\frac{1}{3}\left(\frac{1}{2^{3}} \cdot \frac{1-\frac{1}{2^{n-3}}}{1-\frac{1}{2}}-\frac{1}{2^{2}}\right)+\cdots+\frac{1}{n-1}\left(\frac{1}{2^{n-1}}-\frac{1}{2^{n-2}}\right)+\frac{1}{n}\left(-\frac{1}{2^{n-1}}\right) \\
= & +2\left(\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}\right)+\left(1-\frac{1}{2^{n-1}}\right)+\frac{1}{2}\left(-\frac{1}{2^{n-1}}\right) \\
& +\frac{1}{3}\left(-\frac{1}{2^{n-1}}\right)+\cdots+\frac{1}{n-1}\left(-\frac{1}{2^{n-1}}\right)+\frac{1}{n}\left(-\frac{1}{2^{n-1}}\right) \\
= & 2+2\left(\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}\right) \cdot \frac{1}{2^{n-1}}<2+2\left(\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}\right)
\end{aligned}
$$

构造函数 $f(x)=\ln (1+x)-\frac{x}{1+x}, x>0$

$\therefore f^{\prime}(x)=\frac{x^{2}}{1+x}>0, f(x)$ 在 $(0,+\infty)$ 单调递增

$\therefore f(x)=\ln (1+x)-\frac{x}{1+x}>f(0)=0$

$\therefore \ln (1+x)>\frac{x}{1+x}$ 在 $(0,+\infty)$ 上恒成立, 即 $\frac{x}{1+x}<\ln (1+x)$

令 $x=\frac{1}{n-1}, n \geqslant 2$, 则 $\frac{1}{n}<\ln \left(1+\frac{1}{n-1}\right)$,

从而可得 $\frac{1}{2}<\ln \left(1+\frac{1}{2-1}\right), \frac{1}{3}<\ln \left(1+\frac{1}{3-1}\right), \cdots, \frac{1}{n}<\ln \left(1+\frac{1}{n-1}\right)$,

将以上 $n-1$ 个式子同向相加即得
$\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}<\ln \left(1+\frac{1}{2-1}\right)+\ln \left(1+\frac{1}{3-1}\right)+\cdots+\ln \left(1+\frac{1}{n-1}\right)=\{\}$

$\ln \left(\frac{2}{1} \times \frac{3}{2} \times \cdots \times \frac{n}{n-1}\right)=\ln n$,

故 $S_{n}<2+2\left(\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}\right)<2+2 \ln n$

综上可知, $S_{n}<2+2 \ln n$.
解析：(（1）由题意知: $a_{1}+2 a_{2}+\cdots+n a_{n}=4-\frac{n+2}{2^{n-1}}$

当 $n=3$ 时, $a_{1}+2 a_{2}=4-\frac{2+2}{2^{1}}$;

当 $n=3$ 时, $a_{1}+2 a_{2}+3 a_{3}=4-\frac{3+2}{2^{2}}$;

$3 a_{3}=4-\frac{3+2}{2^{2}}-\left(4-\frac{2+2}{2^{1}}\right)=\frac{3}{4}$

$a_{3}=\frac{1}{4}$

(2) 当 $n=1$ 时, $a_{1}=4-\frac{1+2}{2^{1-1}}=1$;

当 $n \geqslant 2$ 时, 由 $a_{1}+2 a_{2}+\cdots+n a_{n}=4-\frac{n+2}{2^{n-1}}$ 知

$a_{1}+2 a_{2}+\cdots+(n-1) a_{n-1}=4-\frac{n-1+2}{2^{n-2}}$

两式相减得 $n a_{n}=\frac{n+1}{2^{n-2}}-\frac{n+2}{2^{n-1}}=\frac{n}{2^{n-1}}$, 此时 $a_{n}=\frac{1}{2^{n-1}}$.

经检验知 $a_{1}=1$ 也满足 $a_{n}=\frac{1}{2^{n-1}}$. 故数列 $\left\{a_{n}\right\}$ 是以 1 为首项, $\frac{1}{2}$ 为公比的公比数列,

故 $T_{n}=\frac{1 \times\left[1-\left(\frac{1}{2}\right)^{n}\right]}{1-\frac{1}{2}}=2-\frac{1}{2^{n-1}}$.

(3) 由（1）(2) 知, $b_{1}=a_{1}=1$.

当 $n \geqslant 2$ 时, $b_{n}=\frac{T_{n-1}}{n}+\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}\right) a_{n}=\frac{2-\frac{1}{2^{n-2}}}{n}+\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}\right) \cdot \frac{1}{2^{n-1}}$ $=\frac{2}{n}+\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n-1}-\frac{1}{n}\right) \cdot \frac{1}{2^{n-1}}$.

当 $n=1$ 时, $S_{1}=1<2+2 \ln 1=2$, 成立;

当 $n \geqslant 2$ 时,

$$
\begin{aligned}
S_{n}= & +\left[\frac{2}{2}+\left(1-\frac{1}{2}\right) \cdot \frac{1}{2^{1}}\right]+\left[\frac{2}{3}+\left(1+\frac{1}{2}-\frac{1}{3}\right) \cdot \frac{1}{2^{2}}\right]+\cdots \\
& +\left[\frac{2}{n}+\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n-1}-\frac{1}{n}\right) \cdot \frac{1}{2^{n-1}}\right] \\
= & 1+2\left(\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}\right)+\left(\frac{1}{2}+\frac{1}{2^{2}}+\cdots+\frac{1}{2^{n-1}}\right)+\frac{1}{2}\left(\frac{1}{2^{2}}+\frac{1}{2^{3}}+\cdots+\frac{1}{2^{n-1}}-\frac{1}{2}\right) \\
& +\frac{1}{3}\left(\frac{1}{2^{3}}+\frac{1}{2^{4}}+\cdots+\frac{1}{2^{n-1}}-\frac{1}{2^{2}}\right)+\cdots+\frac{1}{n-1}\left(\frac{1}{2^{n-1}}-\frac{1}{2^{n-2}}\right)+\frac{1}{n}\left(-\frac{1}{2^{n-1}}\right) \\
= & 1+2\left(\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}\right)+\left(1-\frac{1}{2^{n-1}}\right)+\frac{1}{2}\left(\frac{1}{2^{2}} \cdot \frac{1-\frac{1}{2^{n-2}}}{1-\frac{1}{2}}-\frac{1}{2}\right) \\
& +\frac{1}{3}\left(\frac{1}{2^{3}} \cdot \frac{1-\frac{1}{2^{n-3}}}{1-\frac{1}{2}}-\frac{1}{2^{2}}\right)+\cdots+\frac{1}{n-1}\left(\frac{1}{2^{n-1}}-\frac{1}{2^{n-2}}\right)+\frac{1}{n}\left(-\frac{1}{2^{n-1}}\right) \\
= & +2\left(\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}\right)+\left(1-\frac{1}{2^{n-1}}\right)+\frac{1}{2}\left(-\frac{1}{2^{n-1}}\right) \\
& +\frac{1}{3}\left(-\frac{1}{2^{n-1}}\right)+\cdots+\frac{1}{n-1}\left(-\frac{1}{2^{n-1}}\right)+\frac{1}{n}\left(-\frac{1}{2^{n-1}}\right) \\
= & 2+2\left(\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}\right) \cdot \frac{1}{2^{n-1}}<2+2\left(\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}\right)
\end{aligned}
$$

构造函数 $f(x)=\ln (1+x)-\frac{x}{1+x}, x>0$

$\therefore f^{\prime}(x)=\frac{x^{2}}{1+x}>0, f(x)$ 在 $(0,+\infty)$ 单调递增

$\therefore f(x)=\ln (1+x)-\frac{x}{1+x}>f(0)=0$

$\therefore \ln (1+x)>\frac{x}{1+x}$ 在 $(0,+\infty)$ 上恒成立, 即 $\frac{x}{1+x}<\ln (1+x)$

令 $x=\frac{1}{n-1}, n \geqslant 2$, 则 $\frac{1}{n}<\ln \left(1+\frac{1}{n-1}\right)$,

从而可得 $\frac{1}{2}<\ln \left(1+\frac{1}{2-1}\right), \frac{1}{3}<\ln \left(1+\frac{1}{3-1}\right), \cdots, \frac{1}{n}<\ln \left(1+\frac{1}{n-1}\right)$,

将以上 $n-1$ 个式子同向相加即得
$\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}<\ln \left(1+\frac{1}{2-1}\right)+\ln \left(1+\frac{1}{3-1}\right)+\cdots+\ln \left(1+\frac{1}{n-1}\right)=\{\}$

$\ln \left(\frac{2}{1} \times \frac{3}{2} \times \cdots \times \frac{n}{n-1}\right)=\ln n$,

故 $S_{n}<2+2\left(\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}\right)<2+2 \ln n$

综上可知, $S_{n}<2+2 \ln n$.