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Worked solution

Take the limit of a given partial sum

Given that ∑n=1N1n(n+1)(n+2)=14−12(N+1)+12(N+2)\displaystyle\sum_{n=1}^{N}\frac1{n(n+1)(n+2)}=\frac14-\frac1{2(N+1)}+\frac1{2(N+2)}, explain whether the series is convergent or divergent, and state the sum to infinity if it is convergent.

CJC · Sequences and Series · Example 26

Answer

The series is convergent; its sum to infinity is 14\frac14.

Compare your reasoning

Take the limits of the two remainder terms

lim⁡N→∞12(N+1)=0,lim⁡N→∞12(N+2)=0\lim_{N\to\infty}\frac1{2(N+1)}=0,\qquad\lim_{N\to\infty}\frac1{2(N+2)}=0

Use the given finite identity. Both denominators grow without bound as N tends to infinity.

Identify the finite limit of the partial sums

lim⁡N→∞∑n=1N1n(n+1)(n+2)=14−0+0=14\begin{aligned}&\lim_{N\to\infty}\sum_{n=1}^{N}\frac1{n(n+1)(n+2)}\\ & =\frac14-0+0\\ & =\frac14\end{aligned}

The Teacher solution concludes convergence and gives the limiting sum.

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