已知 un=2n−1n(n−1)u_n = \frac{2n-1}{n(n-1)}，其中 n∈Z+n \in \mathbb{Z}^+。 证明 un−un+1=2(n+1)(n−1)u_n - u_{n+1} = \frac{2}{(n+1)(n-1)}。 已知 un=2n−1n(n−1)u_n = \frac{2n-1}{n(n-1)}，其中 n∈Z+n \in \mathbb{Z}^+。 证明 un−un+1=2(n+1)(n−1)u_n - u_{n+1} = \frac{2}{(n+1)(n-1)}。

  1. (i)

    由此，求 ∑r=2N1(r+1)(r−1)\sum_{r=2}^{N} \frac{1}{(r+1)(r-1)}，用含 NN 的式子表示你的答案。

    由此，求 ∑r=2N1(r+1)(r−1)\sum_{r=2}^{N} \frac{1}{(r+1)(r-1)}，用含 NN 的式子表示你的答案。

  2. (ii)

    给出第 (i) 部分中的级数收敛的理由，并求其无穷级数之和。

    给出第 (i) 部分中的级数收敛的理由，并求其无穷级数之和。

  3. (iii)

    利用第 (i) 部分中的结果，证明 ∑r=2N−11(r+2)(r)=512−2N+12N(N+1)\sum_{r=2}^{N-1} \frac{1}{(r+2)(r)} = \frac{5}{12} - \frac{2N+1}{2N(N+1)}。

    利用第 (i) 部分中的结果，证明 ∑r=2N−11(r+2)(r)=512−2N+12N(N+1)\sum_{r=2}^{N-1} \frac{1}{(r+2)(r)} = \frac{5}{12} - \frac{2N+1}{2N(N+1)}。

裂项相消求和与差分级数收敛与无穷求和通项公式English
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完整解法

7(i)

1. Relate the general term of the series to the identity from the previous part.

From the previous part, 2(r+1)(r−1)=ur−ur+1\frac{2}{(r+1)(r-1)} = u_r - u_{r+1}.

Therefore, 1(r+1)(r−1)=12(ur−ur+1)\frac{1}{(r+1)(r-1)} = \frac{1}{2}(u_r - u_{r+1}).

2. Apply the method of differences to evaluate the sum.

∑r=2N1(r+1)(r−1)=12∑r=2N(ur−ur+1)\sum_{r=2}^{N} \frac{1}{(r+1)(r-1)} = \frac{1}{2} \sum_{r=2}^{N} (u_r - u_{r+1})

=12[(u2−u3)+(u3−u4)+⋯+(uN−uN+1)]= \frac{1}{2} [ (u_2 - u_3) + (u_3 - u_4) + \dots + (u_N - u_{N+1}) ]

=12(u2−uN+1)= \frac{1}{2} (u_2 - u_{N+1})

3. Substitute the expression for $u_n$ to obtain the final result in terms of $N$.

Since un=2n−1n(n−1)u_n = \frac{2n-1}{n(n-1)}, we evaluate u2u_2 and uN+1u_{N+1}:

u2=2(2)−12(1)=32u_2 = \frac{2(2)-1}{2(1)} = \frac{3}{2}

uN+1=2(N+1)−1(N+1)(N)=2N+1N(N+1)u_{N+1} = \frac{2(N+1)-1}{(N+1)(N)} = \frac{2N+1}{N(N+1)}

Sum=12(32−2N+1N(N+1))\text{Sum} = \frac{1}{2} \left( \frac{3}{2} - \frac{2N+1}{N(N+1)} \right)

=34−2N+12N(N+1)= \frac{3}{4} - \frac{2N+1}{2N(N+1)}

答案

34−2N+12N(N+1)\frac{3}{4} - \frac{2N+1}{2N(N+1)}

7(ii)

1. Establish the expression for the sum of the first N terms (from part i) to analyse its limit.

∑r=2N1(r+1)(r−1)=12∑r=2N(ur−ur+1)=12(u2−uN+1)\sum_{r=2}^{N} \frac{1}{(r+1)(r-1)} = \frac{1}{2} \sum_{r=2}^{N} (u_r - u_{r+1}) = \frac{1}{2} (u_2 - u_{N+1})

Since un=2n−1n(n−1)u_n = \frac{2n-1}{n(n-1)}, u2=32u_2 = \frac{3}{2} and uN+1=2N+1(N+1)Nu_{N+1} = \frac{2N+1}{(N+1)N}.

SN=12(32−2N+1N(N+1))=34−2N+12N(N+1)S_N = \frac{1}{2} \left( \frac{3}{2} - \frac{2N+1}{N(N+1)} \right) = \frac{3}{4} - \frac{2N+1}{2N(N+1)}

2. Provide a valid reason for the convergence of the series.

As N→∞N \to \infty, the denominator of the fractional term 2N+12N(N+1)\frac{2N+1}{2N(N+1)} is a quadratic in NN while the numerator is linear, so 2N+12N(N+1)→0\frac{2N+1}{2N(N+1)} \to 0.

Since the sequence of partial sums SNS_N approaches a finite limit, the series is convergent.

3. Calculate the sum to infinity.

S∞=lim⁡N→∞(34−2N+12N(N+1))=34−0=34S_\infty = \lim_{N \to \infty} \left( \frac{3}{4} - \frac{2N+1}{2N(N+1)} \right) = \frac{3}{4} - 0 = \frac{3}{4}

答案

Reason: As N→∞N \to \infty, 2N+12N(N+1)→0\frac{2N+1}{2N(N+1)} \to 0, so the sequence of partial sums approaches a finite limit.
Sum to infinity: 34\frac{3}{4}

7(iii)

1. Evaluate the sum in part (i) as it is required for this part.

∑r=2N1(r+1)(r−1)=12∑r=2N(ur−ur+1)\sum_{r=2}^{N} \frac{1}{(r+1)(r-1)} = \frac{1}{2} \sum_{r=2}^{N} (u_r - u_{r+1})
=12(u2−uN+1)= \frac{1}{2} (u_2 - u_{N+1})
=12(2(2)−12(1)−2(N+1)−1(N+1)N)= \frac{1}{2} \left( \frac{2(2)-1}{2(1)} - \frac{2(N+1)-1}{(N+1)N} \right)
=12(32−2N+1N(N+1))=34−2N+12N(N+1)= \frac{1}{2} \left( \frac{3}{2} - \frac{2N+1}{N(N+1)} \right) = \frac{3}{4} - \frac{2N+1}{2N(N+1)}

2. Relate the target series to the series from part (i) by substituting the index.

Let k=r+1k = r+1. When r=2r=2, k=3k=3. When r=N−1r=N-1, k=Nk=N.
∑r=2N−11(r+2)(r)=∑k=3N1(k+1)(k−1)\sum_{r=2}^{N-1} \frac{1}{(r+2)(r)} = \sum_{k=3}^{N} \frac{1}{(k+1)(k-1)}

3. Express the shifted series using the result from part (i) and subtract the missing first term.

∑k=3N1(k+1)(k−1)=∑k=2N1(k+1)(k−1)−1(2+1)(2−1)\sum_{k=3}^{N} \frac{1}{(k+1)(k-1)} = \sum_{k=2}^{N} \frac{1}{(k+1)(k-1)} - \frac{1}{(2+1)(2-1)}
=(34−2N+12N(N+1))−13= \left( \frac{3}{4} - \frac{2N+1}{2N(N+1)} \right) - \frac{1}{3}

4. Simplify the constants to obtain the final expression.

=(34−13)−2N+12N(N+1)= \left( \frac{3}{4} - \frac{1}{3} \right) - \frac{2N+1}{2N(N+1)}
=9−412−2N+12N(N+1)= \frac{9-4}{12} - \frac{2N+1}{2N(N+1)}
=512−2N+12N(N+1)(Shown)= \frac{5}{12} - \frac{2N+1}{2N(N+1)} \quad \text{(Shown)}

答案

\frac{5}{12} - \frac{2N+1}{2N(N+1)} \quad \text{(Shown)}

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