Telescoping logs and a rational GP ratio

(a) A sequence u2,u3,u4,…u_2,u_3,u_4,\ldots is such that un=ln⁡(n−1n+1)u_n=\ln(\frac{n-1}{n+1}), n≥2n\ge2. Find in terms of NN an expression for SN=u2+u3+⋯+uN−1+uNS_N=u_2+u_3+\cdots+u_{N-1}+u_N, leaving your answer as ln⁡a−f(N)\ln a-f(N), where aa is positive. marks

(b)(i) Find the set of xx for which ∑n=1∞(3xx+1)n\sum_{n=1}^{\infty}(\frac{3x}{x+1})^n has a finite sum. marks

(ii) Given x=0.25x=0.25, find its value. marks

[2023/MI/Prelim/II/3]

Total marks: 8

Solution

(a)

Telescoping logarithms

un=ln⁡(n−1n+1)u_n=\ln\left(\frac{n-1}{n+1}\right) for n≥2n\ge2. Find SN=u2+u3+⋯+uNS_N=u_2+u_3+\cdots+u_N as ln⁡a−f(N)\ln a-f(N).

Explore this idea

Hint

  1. Write out several terms and watch the fractions cancel.

  2. Write the sum as one logarithm of a product.

Answer · (a)

ln⁡2−ln⁡[N(N+1)]\ln2-\ln[N(N+1)]; a=2a=2, f(N)=ln⁡[N(N+1)]f(N)=\ln[N(N+1)].

Explore the idea

Negative bars that add up to a tidy closed form

Every term ln⁡r−1r+1\ln\frac{r-1}{r+1} is negative, so the bars hang below the axis. Compare the total with the closed form.

Upper limitr = 2 to 6
Try an example
Σ ln((r−1)/(r+1)), r = 2 to 6
-3.045
Number of terms
5
ln 2 − ln N(N+1)
-3.045
matches the direct total
Key idea

The terms shrink toward 00, but their sum has no lower bound, since −ln⁡[N(N+1)]→−∞-\ln[N(N+1)]\to-\infty.

Ideas to try
  1. At N=3N=3 the total is ln⁡13+ln⁡24=ln⁡16\ln\frac13+\ln\frac24=\ln\frac16, and the formula gives ln⁡2−ln⁡12=ln⁡16\ln2-\ln12=\ln\frac16.
  2. Switch to the running total. It keeps falling, because each term is negative.
(b)(i)

Condition for a finite GP sum

Find the set of xx for which ∑n=1∞(3xx+1)n\sum_{n=1}^\infty\left(\frac{3x}{x+1}\right)^n has a finite sum.

Explore this idea

Hint

  1. Identify the common ratio.

  2. Solve ∣r∣<1|r|<1 algebraically or by graph.

Answer · (b)(i)

−1/4<x<1/2-1/4<x<1/2.

Explore the idea

Where is the ratio smaller than 1 in size?

Set k = 1 and read off where the curve y = |3x/(x+1)| lies below the line.

Height of the horizontal linek = 1
Try an example
Test boundary heights

Buttons use full-precision boundary values; numbers shown here are rounded.

Number of distinct intersections
2

At k=1k=1 the crossings are x=−1/4x=-1/4 and x=1/2x=1/2. The modulus is below 1 strictly between them.

All listed intersections are within the original drawing window. A touching point counts once; hollow endpoints are excluded.

Coordinates are rounded to 6 significant figures. Inspect a row to highlight that intersection and bring it into view.
Pointxy
-0.251
0.51
  • Modulus of the common ratio
  • Y = k
  • Intersections
  • Intersection selected in the coordinate table
Key idea

A geometric series converges exactly where the graph of ∣r∣|r| is below 11. That is the open interval between the two crossings.

Ideas to try
Place the line at k=1k=1 and find the two crossing points.
(b)(ii)

Evaluate the sum to infinity

Given x=0.25x=0.25, find the value of the sum.

Explore this idea

Hint

  1. Compute the ratio at x=0.25x=0.25.

  2. Use a1−r\frac a{1-r} with a=ra=r.

Answer · (b)(ii)

3/23/2.

Explore the idea

Powers of 3/5 adding up to 3/2

The bars shrink geometrically. The running total creeps toward 1.51.5 but never reaches it.

Upper limitr = 1 to 4
Try an example
Σ (3/5)^r, r = 1 to 4
1.306
Number of terms
4
1.5(1 − 0.6^N)
1.306
matches the direct total
Key idea

Adding more terms brings the total as close as you like to 32\frac32.

Ideas to try
Open the running total and move NN to 1414. The total is within a hair of 1.51.5.
Takeaway

Collapse, then bound

Logarithms of ratios telescope. For a geometric series, a finite sum exists exactly when the ratio's modulus is below 11.