A factorial tail after reindexing

It is given that ∑r=1nr2+3r+1(r+2)!=32−1(n+1)!−1(n+2)!\sum_{r=1}^n\frac{r^2+3r+1}{(r+2)!}=\frac32-\frac1{(n+1)!}-\frac1{(n+2)!}.

(a) Explain why its infinite series converges, and state the limit. marks

(b) Find exactly ∑r=6∞r2−r−1r!\sum_{r=6}^{\infty}\frac{r^2-r-1}{r!}. marks

[modified 2024/EJC/Prelim/I/6]

Total marks: 5

Solution

(a)

Convergence and limit

Given ∑r=1nr2+3r+1(r+2)!=32−1(n+1)!−1(n+2)!\sum_{r=1}^n\frac{r^2+3r+1}{(r+2)!}=\frac32-\frac1{(n+1)!}-\frac1{(n+2)!}, explain why the infinite series converges, and state its limit.

Explore this idea

Hint

  1. Take n→∞n\to\infty in the formula.

  2. Say that the limit is finite.

Answer · (a)

Converges to 3/23/2: both factorial remainders tend to zero.

Explore the idea

Terms vanishing almost instantly

The bars shrink faster than any geometric series. The running total is within a hair of 1.51.5 after only a few terms.

Upper limitr = 1 to 3
Try an example
Σ (r²+3r+1)/(r+2)!, r = 1 to 3
1.45
Number of terms
3
3/2 − 1/(n+1)! − 1/(n+2)!
1.45
matches the direct total
Key idea

Factorial denominators give such fast decay that a handful of terms already pins the sum to several decimal places.

Ideas to try
Open the running total at n=4n=4. How far from 1.51.5 is it?
(b)

Reindex then subtract

Find exactly ∑r=6∞r2−r−1r!\sum_{r=6}^\infty\frac{r^2-r-1}{r!}.

Explore this idea

Hint

  1. Replace rr by r+2r+2 so the term matches part (a).

  2. Write the tail as the whole series minus a head.

Answer · (b)

1/201/20.

Explore the idea

The same series, labelled differently

The bars are r2−r−1r!\frac{r^2-r-1}{r!} for r≥3r\ge3. Selecting rr from 66 gives the tail; the closed form uses the shift s=r−2s=r-2.

Limits of the sumr = 6 to 13
Try an example
Σ (r²−r−1)/r!, r = 6 to 13
0.05
Number of terms
8
S(upper−2) − S(lower −3)
0.05
matches the direct total
Key idea

A change of variable s=r−2s=r-2 turns an unfamiliar sum into the one that was given, with the limits shifted to match.

Ideas to try
  1. With the block 66 to 1313 the total is 0.05=1200.05=\frac1{20} to many decimal places.
  2. Move the lower limit to 33: the same formula with S0=0S_0=0 now counts every term.
Takeaway

Match the summand first

To reuse a given result, rewrite the new sum so its general term is identical, then adjust the limits by the same shift.