A telescoping increment

u_1,u_2,u_3,\ldots

[2011/DHS/Prelim/II/4b(i),(ii)]

Total marks: 3

Solution

(i)

Iterate a recurrence

$u_1=\frac32$ and $u_{n+1}=u_n+\frac1{(n+1)(n+2)}$ for $n\in\mathbb Z^+$ . Find $u_2,u_3,u_4$ .

Explore this idea

Hint

Substitute $n=1,2,3$ in turn.
Use each new term as the next input; keep every answer as an exact fraction.

Answer · (i)

u_2=5/3,u_3=7/4,u_4=9/5

Explore the idea

Each term adds one small slice

The bars are the increments $\frac1{(r+1)(r+2)}$ . Switch to the running total to see $u_n-u_1$ build up. The total can never pass $\frac12$ .

Upper limitr = 1 to 3

Try an example

Σ 1/((r+1)(r+2)), r = 1 to 3

0.3

Number of terms

½ − 1/(upper+2)

0.3

matches the direct total

Key idea

Adding the increments from $r=1$ to $r=n-1$ takes $u_1$ to $u_n$ , so every printed term is $\frac32$ plus a running total.

Ideas to try

Set the upper limit to $3$ and read the total: $u_4=\frac32+(\text{total})=\frac95$ .
Slide the upper limit to the right. The slices shrink so quickly that the total stalls near $0.5$ .

(ii)

Conjecture a closed form

Form a conjecture for $u_n$ in terms of $n$ .

Explore this idea

Hint

Write $u_n$ as a fraction in $n$ using the pattern in part (i).
Test it, then explain it by splitting the increment into partial fractions.

Answer · (ii)

u_n=2-1/(n+1)

Explore the idea

The terms rise towards 2

The solid curve is the recurrence's terms; the dashed curve is the conjectured $2-\frac1{n+1}$ . They coincide at every integer.

Indexn = 4

Try an example

term uₙ at n = 4

1.8

2 − 1/(n+1) at n = 4

1.8

Key idea

A telescoping increment makes $u_n$ equal to a constant minus a single shrinking fraction, so the terms climb to $2$ without ever reaching it.

Ideas to try

Slide

n

upward. The gap to the red line is exactly

\frac1{n+1}

Takeaway

Cancel before you add

When the increment is a product of consecutive linear factors, partial fractions turn a long sum into just two surviving terms.

A telescoping increment

Solution

Iterate a recurrence

Second term

Third term

Fourth term

Each term adds one small slice

Conjecture a closed form

Read the pattern

Test the rule against the recurrence

See why it works

The terms rise towards 2

Cancel before you add