已知 sumr=1nfrac1r(r+1)=1−frac1n+1\\sum_{r=1}^{n} \\frac{1}{r(r + 1)} = 1 - \\frac{1}{n + 1}。

  1. (a)

    求 sumr=5n+1frac1r(r+1)\\sum_{r=5}^{n+1} \\frac{1}{r(r + 1)}。

  2. (b)

    给出 (a) 部分中的级数收敛的原因，并说明 sumr=5inftyfrac1r(r+1)\\sum_{r=5}^{\\infty} \\frac{1}{r(r + 1)} 的值。

裂项相消求和与差分级数收敛与无穷求和通项公式English
先不看答案，让 AI 诊断你卡在哪里？

它不会直接告诉你答案，而是问你卡在哪一步看看 AI 是怎么陪你一步步弄懂这道题的

完整解法

3(a)

1. Express the sum starting from $r=5$ as the difference of two series starting from $r=1$.

∑r=5n+11r(r+1)=∑r=1n+11r(r+1)−∑r=141r(r+1)\sum_{r=5}^{n+1} \frac{1}{r(r + 1)} = \sum_{r=1}^{n+1} \frac{1}{r(r + 1)} - \sum_{r=1}^{4} \frac{1}{r(r + 1)}

2. Evaluate both series using the given identity.

Substitute N=n+1N = n+1 and N=4N = 4 into the given formula ∑r=1N1r(r+1)=1−1N+1\sum_{r=1}^{N} \frac{1}{r(r + 1)} = 1 - \frac{1}{N + 1}:

∑r=1n+11r(r+1)=1−1(n+1)+1=1−1n+2\sum_{r=1}^{n+1} \frac{1}{r(r + 1)} = 1 - \frac{1}{(n+1) + 1} = 1 - \frac{1}{n+2}

∑r=141r(r+1)=1−14+1=1−15=45\sum_{r=1}^{4} \frac{1}{r(r + 1)} = 1 - \frac{1}{4 + 1} = 1 - \frac{1}{5} = \frac{4}{5}

3. Substitute the evaluated sums back and simplify the resulting expression.

∑r=5n+11r(r+1)=(1−1n+2)−45\sum_{r=5}^{n+1} \frac{1}{r(r + 1)} = \left( 1 - \frac{1}{n+2} \right) - \frac{4}{5}
=15−1n+2= \frac{1}{5} - \frac{1}{n+2}
=n+2−55(n+2)= \frac{n + 2 - 5}{5(n + 2)}
=n−35(n+2)= \frac{n - 3}{5(n + 2)}

答案

\frac{n - 3}{5(n + 2)}

3(b)

1. Provide the reason for the convergence of the series based on the expression from part (a)

From part (a), we know that ∑r=5n+11r(r+1)=15−1n+2\sum_{r=5}^{n+1} \frac{1}{r(r + 1)} = \frac{1}{5} - \frac{1}{n+2}.
As n→∞n \to \infty, the term 1n+2→0\frac{1}{n+2} \to 0.
Since the limit exists and is finite, the series is convergent.

2. State the value of the sum to infinity

∑r=5∞1r(r+1)=15−0=15\sum_{r=5}^{\infty} \frac{1}{r(r + 1)} = \frac{1}{5} - 0 = \frac{1}{5}

答案

Reason: As n→∞n \to \infty, 1n+2→0\frac{1}{n+2} \to 0.
Value: 15\frac{1}{5}

AI 分析，可能出错

用 AI 批改，圈出写错的那一步，并自动将你的薄弱点存入错题本，形成你的错因库。

同一知识点的其他题

  1. 1已知 f(r)=r(r−1)!f(r)=\dfrac{r}{(r-1)!}，证明 f(r−1)−f(r)=r2−3r+1(r−1)!f(r-1)-f(r)=\dfrac{r^2-3r+1}{(r-1)!}。 由此求 ∑r=2nr2−3r+1(r−1)!\displaystyle\sum_{r=2}^{n} \frac{r^2-3r+1}{(r-1)!}…
  2. 2已知数列{an}\{a_{n}\}中，a1=76a_{1}=\dfrac{7}{6}，an2−an+1=an+1(n∈N∗)a_{n}^{2}-a_{n}+1=a_{n+1}(n\in N^{*})，SnS_{n}是{1an}\{\dfrac{1}{a_{n}}\}的前…
  3. 3已知 f(r)=r(r−1)!f(r)=\dfrac{r}{(r-1)!}，证明 f(r−1)−f(r)=r2−3r+1(r−1)!f(r-1)-f(r)=\dfrac{r^2-3r+1}{(r-1)!}。 由此求 ∑r=2nr2−3r+1(r−1)!\displaystyle\sum_{r=2}^n \dfrac{r^2-3r+1}{(r-1)!}…
  4. 4证明 32r+1−42r+3+12r+5=8r+k(2r+1)(2r+3)(2r+5)\frac{3}{2r+1}-\frac{4}{2r+3}+\frac{1}{2r+5}=\frac{8r+k}{(2r+1)(2r+3)(2r+5)}，其中 kk 是待求常数。 由此求 ∑r=0n2r+7(2r+1)(2r+3)(2r+5)\sum_{r=0}^n \frac{2r+7}{(2r+1)(2r+3)(2r+5)}…
  5. 5用数学归纳法证明，对于所有正整数 nn 以及 x∈Rx\in\mathbb{R}，x≠0, x≠1x\ne 0,\ x\ne 1， 1x−1−1x−1x2−⋯−1xn=1xn(x−1) .\frac{1}{x-1}-\frac{1}{x}-\frac{1}{x^2}-\dots-\frac{1}{x^n}=\frac{1}{x^n(x-1)}\,.…