Alternating terms, even terms and products

A geometric series G is given by e−e1+α+e1+2α−e1+3α+⋯e-e^{1+\alpha}+e^{1+2\alpha}-e^{1+3\alpha}+\cdots, where α\alpha is a negative real number.

(i) Show that G is convergent. marks

(ii) Its sum to infinity is SS. Given that the sum to infinity of the even-numbered terms (the second, fourth, sixth, … terms) is −13S-\frac13S, find α\alpha exactly. marks

(iii) In another series H, each term is the modulus of the corresponding G term. Given α=−0.03\alpha=-0.03, find the greatest nn such that the product of H’s first nn terms exceeds 15000. marks

[2009/IJC/Prelim/I/12]

Total marks: 10

Solution

(i)

Show convergence

Show that GG converges when α<0\alpha<0.

Hint

  1. What does α<0 imply about eαe^\alpha?

Answer · (i)

∣−eα∣<1|-e^\alpha|<1, so converges.

(ii)

Determine α

Find α\alpha when the even-indexed terms sum to −S/3-S/3.

Hint

  1. The even subsequence starts with −e1+α-e^{1+\alpha} and ratio e2αe^{2\alpha}.

Answer · (ii)

α=−ln⁡4\alpha=-\ln4.

(iii)

Bound a product

For α=−0.03\alpha=-0.03, find the greatest nn for which the product of the first nn absolute terms exceeds 15000.

Explore this idea

Hint

  1. For a product of exponentials, sum the exponents first.

  2. Do not assume adding more factors always increases the product.

Answer · (iii)

n=56n=56.

Explore the idea

Why the product eventually falls

The vertical axis is ln(product). Later factors below 1 eventually reduce the product.

Factor countn = 56
Try an example
ln Pₙ at n = 56
9.8
Key idea

A product need not grow when a new factor is added. Here the last qualifying integer is 56.

Ideas to try
Compare the values on either side of the boundary. What makes one count qualify?
Takeaway

Separate alternating sums from products

There are n exponents 1,1+α,…,1+(n−1)α1,1+α,\ldots,1+(n-1)α, forming an AP.