An error bound and a shifted rational sum

The sum ∑r=1n4(2r+3)(2r+5)\sum_{r=1}^n\frac4{(2r+3)(2r+5)} is denoted SnS_n. Given Sn=25−22n+5S_n=\frac25-\frac2{2n+5},

(i) find the smallest nn for which SnS_n is within 10−310^{-3} of the sum to infinity marks;

(ii) find ∑r=102n24r2−1\sum_{r=10}^{2n}\frac2{4r^2-1} in terms of nn. marks

[modified 2020/ASRJC/Prelim/I/5]

Total marks: 6

Solution

(i)

Smallest n within a tolerance

Sn=∑r=1n4(2r+3)(2r+5)=25−22n+5S_n=\sum_{r=1}^n\frac4{(2r+3)(2r+5)}=\frac25-\frac2{2n+5}. Find the smallest nn for which SnS_n is within 10−310^{-3} of the sum to infinity.

Explore this idea

Hint

  1. Find S∞S_\infty.

  2. Solve ∣Sn−S∞∣<10−3|S_n-S_\infty|<10^{-3}, remembering that "within" means strictly less.

Answer · (i)

n=998n=998.

Explore the idea

The gap shrinks below 0.001

The gap 22n+5\frac2{2n+5} against the tolerance. Use the presets to compare the two terms either side of the threshold.

Number of termsn = 998
Try an example
gap 2/(2n+5) at n = 998
0.0009995
Key idea

A strict inequality with a half-integer boundary 997.5997.5 sends the answer up to the next integer, 998998.

Ideas to try
Compare the readouts at 997997 and 998998. One gap is just above 0.0010.001 and the other just below.
(ii)

Reindex a rational sum

Find ∑r=102n24r2−1\sum_{r=10}^{2n}\frac2{4r^2-1} in terms of nn.

Explore this idea

Hint

  1. Factorise 4r2−1=(2r−1)(2r+1)4r^2-1=(2r-1)(2r+1).

  2. Shift r→r+2r\to r+2 to match (2r+3)(2r+5)(2r+3)(2r+5).

  3. Subtract two sums that start at 11.

Answer · (ii)

1/19−1/(4n+1)1/19-1/(4n+1).

n∈Zn\in\mathbb Z, n≥5n\ge5.

Explore the idea

A sum that telescopes by itself

Each term 24r2−1\frac2{4r^2-1} equals 12r−1−12r+1\frac1{2r-1}-\frac1{2r+1}, so the sum from aa to bb is 12a−1−12b+1\frac1{2a-1}-\frac1{2b+1}.

Limits of the sumr = 10 to 14
Try an example
Σ 2/(4r²−1), r = 10 to 14
0.018
Number of terms
5
1/(2a−1) − 1/(2b+1)
0.018
matches the direct total
Key idea

With 2n2n as the upper limit, 2b+1=4n+12b+1=4n+1, which is where the answer's denominator comes from.

Ideas to try
  1. Set the block 1010 to 1414 (like n=7n=7). The total is 119−129\frac1{19}-\frac1{29}.
  2. Check that 119−14n+1\frac1{19}-\frac1{4n+1} with n=7n=7 gives the same number.
Takeaway

Tolerance and telescoping

A strict tolerance gives a boundary value that must be rounded up. A rational summand with consecutive odd factors telescopes.