数列 $u_1, u_2, u_3, \dots$ 满足 $u_1 = \dfrac{e}{10}$ 且 对于所有 $n \ge 2$，$u_n - u_{n-1} = \dfrac{e}{2^{n-1}}$。 使用裂项消去法证明： $u_n = \dfrac{e}{10}\left[11 - 10\left(\dfrac{1}{2}\right)^{n-1}\right].$ 指出 $u_n$ 是否为收敛数列，并说明理由。 数列 $a_1, a_2, a_3, \dots$ 定义为 $a_r = \dfrac{1}{1+2+3+\dots+ r},$ 其中 $r$ 为正整数。 另一个数列 $b_1, b_2, b_3, \dots$ 定义为 $b_n = \dfrac{1}{4}\left(\sum_{r=1}^{n} a_r\right),$ 其中 $n$ 为正整数。 求项 $b_1, b_2$ 和 $b_3$。由此对 $b_n$ 作出猜想，并将你的答案表示为 $\dfrac{n}{f(n)}$ 的形式。 使用数学归纳法证明你的猜想。

1. Evaluate the limit of $u_n$ as $n \to \infty$ to conclude on convergence.

As $n \to \infty$, $\left(\frac{1}{2}\right)^{n-1} \to 0$.
Therefore, $\lim_{n \to \infty} u_n = \frac{e}{10}(11 - 0) = \frac{11e}{10}$.
Since $u_n$ approaches a finite value as $n \to \infty$, it is a convergent sequence.

1. Evaluate the first three terms of sequence $a_r$ and use them to calculate $b_1, b_2$ and $b_3$.

$a_1 = \dfrac{1}{1} = 1$
$a_2 = \dfrac{1}{1+2} = \dfrac{1}{3}$
$a_3 = \dfrac{1}{1+2+3} = \dfrac{1}{6}$

$b_1 = \dfrac{1}{4}(a_1) = \dfrac{1}{4}(1) = \dfrac{1}{4}$

$b_2 = \dfrac{1}{4}(a_1 + a_2) = \dfrac{1}{4}\left(1 + \dfrac{1}{3}\right) = \dfrac{1}{4}\left(\dfrac{4}{3}\right) = \dfrac{1}{3}$

$b_3 = \dfrac{1}{4}(a_1 + a_2 + a_3) = \dfrac{1}{4}\left(\dfrac{4}{3} + \dfrac{1}{6}\right) = \dfrac{1}{4}\left(\dfrac{9}{6}\right) = \dfrac{3}{8}$

2. Rewrite $b_1, b_2, b_3$ to match the required form $\frac{n}{f(n)}$ and state the conjecture.

To express $b_n$ in the form $\dfrac{n}{f(n)}$, rewrite the terms such that the numerator is the term number $n$:
$b_1 = \dfrac{1}{4}$
$b_2 = \dfrac{1}{3} = \dfrac{2}{6}$
$b_3 = \dfrac{3}{8}$

The denominators form an arithmetic sequence $4, 6, 8, \dots$, which can be described by $f(n) = 2n + 2$.

Therefore, the conjecture is:
$b_n = \dfrac{n}{2n+2}$

1. State the conjecture derived from the previous part and define the proposition.

From part (b)(i), we found that $a_r = \frac{2}{r(r+1)}$ and conjectured that $b_n = \frac{n}{2(n+1)}$.
Let $P_n$ be the statement that $b_n = \frac{n}{2(n+1)}$ for all $n \in \mathbb{Z}^+$.

2. Verify the base case for $n=1$.

When $n = 1$,
$\text{LHS} = b_1 = \frac{1}{4} a_1 = \frac{1}{4} (1) = \frac{1}{4}$
$\text{RHS} = \frac{1}{2(1+1)} = \frac{1}{4}$
Since $\text{LHS} = \text{RHS}$, $P_1$ is true.

3. State the inductive hypothesis.

Assume that $P_k$ is true for some positive integer $k$, i.e.,
$b_k = \frac{k}{2(k+1)}$

4. Perform the inductive step to show that $P_k \implies P_{k+1}$.

We want to show that $P_{k+1}$ is true, i.e., $b_{k+1} = \frac{k+1}{2(k+2)}$.
Using the definition of the sequence, $b_{k+1} = b_k + \frac{1}{4}a_{k+1}$.
Since $a_{k+1} = \frac{2}{(k+1)(k+2)}$, we have:

Thus, $P_{k+1}$ is true if $P_k$ is true.

5. State the concluding statement of mathematical induction.

Since $P_1$ is true, and $P_k$ is true implies $P_{k+1}$ is true, by the Principle of Mathematical Induction, $P_n$ is true for all positive integers $n$.