Porisma

Worked solution

Shift a rational sum to match a given identity

Given ∑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)}, find ∑n=3N1n(n−1)(n−2)\displaystyle\sum_{n=3}^{N}\frac1{n(n-1)(n-2)} for integer N≥3N\ge3.

CJC · Sequences and Series · Example 17

Answer

14−12(N−1)+12N\frac14-\frac1{2(N-1)}+\frac1{2N}.

Compare your reasoning

Method 1: list the required terms

∑n=3N1n(n−1)(n−2)=13(3−1)(3−2)+14(4−1)(4−2)+15(5−1)(5−2)+⋯+1(N−1)(N−2)(N−3)+1N(N−1)(N−2).\begin{aligned}\sum_{n=3}^{N}\frac1{n(n-1)(n-2)}&=\frac1{3(3-1)(3-2)}+\frac1{4(4-1)(4-2)}\\&\quad+\frac1{5(5-1)(5-2)}+\cdots\\&\quad+\frac1{(N-1)(N-2)(N-3)}+\frac1{N(N-1)(N-2)}.\end{aligned}

Retain the first three and final two displayed terms from the source. This expanded pattern is schematic for a sufficiently long sum; at a small N include each actual index only once.

Reorder the factors to recognise the pattern

11⋅2⋅3+12⋅3⋅4+13⋅4⋅5+⋯+1(N−3)(N−2)(N−1)+1(N−2)(N−1)N=∑n=1N−21n(n+1)(n+2).\begin{aligned}&\frac1{1\cdot2\cdot3}+\frac1{2\cdot3\cdot4}+\frac1{3\cdot4\cdot5}+\cdots\\&\quad+\frac1{(N-3)(N-2)(N-1)}+\frac1{(N-2)(N-1)N}\\&=\sum_{n=1}^{N-2}\frac1{n(n+1)(n+2)}.\end{aligned}

The source says to simplify and reorder terms. The index may be renamed after the list is matched.

Apply the given formula with upper limit N minus two

∑n=3N1n(n−1)(n−2)=14−12[(N−2)+1]+12[(N−2)+2]=14−12(N−1)+12N.\begin{aligned}\sum_{n=3}^{N}\frac1{n(n-1)(n-2)}&=\frac14-\frac1{2[(N-2)+1]}+\frac1{2[(N-2)+2]}\\&=\frac14-\frac1{2(N-1)}+\frac1{2N}.\end{aligned}

Keep the complete source substitution before simplifying.

Method 2: change the counter

r=n−2,n−1=r+1,n=r+2,1n(n−1)(n−2)=1(r+2)(r+1)rr=n-2,\qquad n-1=r+1,\qquad n=r+2,\qquad\frac1{n(n-1)(n-2)}=\frac1{(r+2)(r+1)r}

The illustrated callout uses a different letter to prevent confusion. Its goal is to match the given denominator and start at index 1.

Transform both bounds and reorder the factors

n=3⟹r=3−2=1,n=N⟹r=N−2,∑n=3N1n(n−1)(n−2)=∑r=1N−21(r+2)(r+1)r=∑r=1N−21r(r+1)(r+2).\begin{aligned}n=3&\Longrightarrow r=3-2=1,\\n=N&\Longrightarrow r=N-2,\\\sum_{n=3}^{N}\frac1{n(n-1)(n-2)}&=\sum_{r=1}^{N-2}\frac1{(r+2)(r+1)r}\\&=\sum_{r=1}^{N-2}\frac1{r(r+1)(r+2)}.\end{aligned}

The Teacher continuation supplies both endpoint checks.

Use the given result again

14−12[(N−2)+1]+12[(N−2)+2]=14−12(N−1)+12N\begin{aligned}&\frac14-\frac1{2[(N-2)+1]}\\ &{}+\frac1{2[(N-2)+2]}\\ & =\frac14-\frac1{2(N-1)}+\frac1{2N}\end{aligned}

Both printed methods reach the same expression.

How did this question go?

Common mistakes

Changing the denominator without changing the upper limit would include two extra terms. The same letter can be a dummy counter, but N remains the parameter.