TIFR 2013 problem 35 | Sequence Bounded / Unbounded

Let's discuss a problem and find out whether a sequence is bounded or unbounded from TIFR 2013 Problem 35. Try before reading the solution.

Question: TIFR 2013 problem 35

True/False?

Let \( \{a_n\} \) be any non-constant sequence in \(\mathbb{R}\) such that \( a_{n+1}=\frac{a_n + a_{n+2} }{2} \) for all \(n \ge 1 \). Then \(\{a_n\} \) is unbounded.

Hint:

The given expression is same as \( a_{n+1}-a_n = a_{n+2} -a_{n+1} \).

Discussion:

The distance between two successive terms in the given sequence is constant. It is given by \( |a_{n+1}- a_n| = |a_n - a_{n-1}| = ... = |a_1-a_0| \).

So for the sequence to be non-constant, \( a_1 \ne a_0 \). Because otherwise, the sequence will have distance between any two successive terms zero, which is just another way of saying that the sequence is constant.

There are two cases:

Case 1: \(a_1 > a_0\).

Then \(a_{n+1} > a_n\), that is the sequence is increasing, and not only that, it is an arithmetic progression with common difference \(a_1-a_0 (> 0)\). Therefore, the sequence is unbounded above.

Case 2: \(a_1<a_0\).

Then as in the previous case the sequence this time will become a decreasing sequence, not only that, it is an arithmetic progression with common difference \( < 0 \). Therefore, the sequence is unbounded below.

Remark:

We don't really need to take the two cases. The key point is that the given recurrence relation is that of an arithmetic progression whose common difference is non-zero. Hence the sequence has to be unbounded.

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• TIFR Problem Garden
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