Prove both bounds for Tₙ, for every positive integer n.

01 / Standard#Your turn

\frac1{12}\le T_n<\frac3{16}

Official paper · jm01-2026 · II.3(c) · PDF 4

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1

Every bₙ is positive.

Hint 2

The subtracted remainder in the formula is also positive.

Worked solution

Positive summands make the partial sums increasing, with minimum at n=1.
$T_n\ge T_1=b_1=\frac1{4\cdot1\cdot3}=\frac1{12}$
For every finite positive n the remainder is strictly positive.
$\frac18\left(\frac1{n+1}+\frac1{n+2}\right)>0\implies T_n<\frac3{16}$

The lower bound is attained only at n=1; the upper bound is never attained.

Checks and common pitfalls: A limit need not be an attained maximum of the sequence.

Reasoning checklist · self / teacher assessment

Teaching assessment checklist, independently authored. Use the original paper for official marks.
Positive summands make the partial sums increasing, with minimum at n=1.
$T_n\ge T_1=b_1=\frac1{4\cdot1\cdot3}=\frac1{12}$
For every finite positive n the remainder is strictly positive.
$\frac18\left(\frac1{n+1}+\frac1{n+2}\right)>0\implies T_n<\frac3{16}$

Think first. Reveal a hint when the class is ready.

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