Consider the sequence $(a_n)_{n\geqslant 1}$ defined by $a_1=1/2$ and $2n\cdot a_{n+1}=(n+1)a_n.$[list=a]
[*]Determine the general formula for $a_n.$
[*]Let $b_n=a_1+a_2+\cdots+a_n.$ Prove that $\{b_n\}-\{b_{n+1}\}\neq \{b_{n+1}\}-\{b_{n+2}\}.$
[/list]
1. **Determine the general formula for \(a_n\):**

   We are given the sequence \((a_n)_{n \geq 1}\) defined by \(a_1 = \frac{1}{2}\) and \(2n \cdot a_{n+1} = (n+1) a_n\). We need to find a general formula for \(a_n\).

   Let's calculate the first few terms to identify a pattern:
   \[
   a_1 = \frac{1}{2}
   \]
   For \(n = 1\):
   \[
   2 \cdot 1 \cdot a_2 = (1+1) \cdot a_1 \implies 2a_2 = 2 \cdot \frac{1}{2} \implies a_2 = \frac{1}{2}
   \]
   For \(n = 2\):
   \[
   2 \cdot 2 \cdot a_3 = (2+1) \cdot a_2 \implies 4a_3 = 3 \cdot \frac{1}{2} \implies a_3 = \frac{3}{8}
   \]
   For \(n = 3\):
   \[
   2 \cdot 3 \cdot a_4 = (3+1) \cdot a_3 \implies 6a_4 = 4 \cdot \frac{3}{8} \implies a_4 = \frac{1}{4}
   \]

   We observe that \(a_n\) seems to follow the pattern \(a_n = \frac{n}{2^n}\). We will prove this by induction.

   **Base case:**
   For \(n = 1\), \(a_1 = \frac{1}{2}\), which matches the formula.

   **Inductive step:**
   Assume \(a_k = \frac{k}{2^k}\) for some positive integer \(k\). We need to show that \(a_{k+1} = \frac{k+1}{2^{k+1}}\).

   Using the given recurrence relation:
   \[
   2k \cdot a_{k+1} = (k+1) a_k
   \]
   Substitute the inductive hypothesis \(a_k = \frac{k}{2^k}\):
   \[
   2k \cdot a_{k+1} = (k+1) \cdot \frac{k}{2^k}
   \]
   Solving for \(a_{k+1}\):
   \[
   a_{k+1} = \frac{(k+1) \cdot \frac{k}{2^k}}{2k} = \frac{k+1}{2^{k+1}}
   \]
   Thus, by induction, the general formula for \(a_n\) is:
   \[
   a_n = \frac{n}{2^n}
   \]

2. **Prove that \(\{b_n\} - \{b_{n+1}\} \neq \{b_{n+1}\} - \{b_{n+2}\}\):**

   Let \(b_n = a_1 + a_2 + \cdots + a_n\). Using the formula from part (a), we have:
   \[
   b_n = \sum_{k=1}^n \frac{k}{2^k}
   \]

   We need to find a simplified expression for \(b_n\). Consider the series:
   \[
   S = \sum_{k=1}^\infty \frac{k}{2^k}
   \]
   This is a known series and can be evaluated as follows:
   \[
   S = \sum_{k=1}^\infty \frac{k}{2^k} = 2
   \]

   Therefore, for finite \(n\), we have:
   \[
   b_n = 2 - \sum_{k=n+1}^\infty \frac{k}{2^k}
   \]

   The remaining series can be simplified:
   \[
   \sum_{k=n+1}^\infty \frac{k}{2^k} = \frac{n+1}{2^{n+1}} + \frac{n+2}{2^{n+2}} + \cdots
   \]
   This is a geometric series with the first term \(\frac{n+1}{2^{n+1}}\) and common ratio \(\frac{1}{2}\):
   \[
   \sum_{k=n+1}^\infty \frac{k}{2^k} = \frac{n+1}{2^{n+1}} \left(1 + \frac{1}{2} + \frac{1}{4} + \cdots \right) = \frac{n+1}{2^{n+1}} \cdot 2 = \frac{n+1}{2^n}
   \]

   Thus:
   \[
   b_n = 2 - \frac{n+1}{2^n}
   \]

   The fractional part of \(b_n\) is:
   \[
   \{b_n\} = 1 - \frac{n+1}{2^n}
   \]

   Now, we need to show that:
   \[
   \{b_n\} - \{b_{n+1}\} \neq \{b_{n+1}\} - \{b_{n+2}\}
   \]

   Calculate \(\{b_{n+1}\}\) and \(\{b_{n+2}\}\):
   \[
   \{b_{n+1}\} = 1 - \frac{n+2}{2^{n+1}}
   \]
   \[
   \{b_{n+2}\} = 1 - \frac{n+3}{2^{n+2}}
   \]

   Compute the differences:
   \[
   \{b_n\} - \{b_{n+1}\} = \left(1 - \frac{n+1}{2^n}\right) - \left(1 - \frac{n+2}{2^{n+1}}\right) = \frac{n+2}{2^{n+1}} - \frac{n+1}{2^n} = -\frac{n+1}{2^{n+1}}
   \]
   \[
   \{b_{n+1}\} - \{b_{n+2}\} = \left(1 - \frac{n+2}{2^{n+1}}\right) - \left(1 - \frac{n+3}{2^{n+2}}\right) = \frac{n+3}{2^{n+2}} - \frac{n+2}{2^{n+1}} = -\frac{n+2}{2^{n+2}}
   \]

   Assume for contradiction that these two expressions are equal:
   \[
   -\frac{n+1}{2^{n+1}} = -\frac{n+2}{2^{n+2}}
   \]

   Simplifying:
   \[
   \frac{n+1}{2^{n+1}} = \frac{n+2}{2^{n+2}} \implies 2(n+1) = n+2 \implies 2n + 2 = n + 2 \implies n = 0
   \]

   This is a contradiction since \(n \geq 3\). Therefore, the assumption is false, and we have:
   \[
   \{b_n\} - \{b_{n+1}\} \neq \{b_{n+1}\} - \{b_{n+2}\}
   \]

The final answer is \(\boxed{a_n = \frac{n}{2^n}}\) and \(\{b_n\} - \{b_{n+1}\} \neq \{b_{n+1}\} - \{b_{n+2}\}\).