It is given that ∑r=1n1r(r+1)=1−1n+1\sum_{r=1}^{n} \frac{1}{r(r + 1)} = 1 - \frac{1}{n + 1}.

  1. (a)

    Find ∑r=5n+11r(r+1)\sum_{r=5}^{n+1} \frac{1}{r(r + 1)}.

  2. (b)

    Give a reason why the series in part (a) is convergent and state the value of ∑r=5∞1r(r+1)\sum_{r=5}^{\infty} \frac{1}{r(r + 1)}.

Sum difference of seriesConvergence sum to infinityFormula nth term中文
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Worked solution

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)}

Answer

\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}

Answer

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

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