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Arithmetic Sequence of reciprocals | ISI subjective 2015

This is Problem number 7 from the ISI Subjective Entrance Exam based on the Arithmetic Sequence of reciprocals. Try to solve the problem.

Let m_1, m_2 , ... , m_k be k positive numbers such that their reciprocals are in A.P. Show that k< m_1 + 2 . Also find such a sequence for positive integer k > 0.

......

Teacher: This is a simple application of Arithmetic Mean - Harmonic Mean Inequality. First notice that if k \le 2 we have nothing to prove as m_1 \ge 1 , m_1 +2 is definitely greater than k.

So we are interested in the cases where k \ge 3
Now apply the A.M. - H.M. inequality

Student: Ok. \displaystyle { \frac {k}{\frac{1}{m_1} + \frac {1}{m_2} + ... + \frac{1}{m_k}} \ge \frac{m_1 + m_2 + ... + m_k}{k} }

Now we already know that \displaystyle { \frac{1}{m_1} , \frac {1}{m_2} , ... , \frac{1}{m_k} } are in A.P. So sum of these k terms in A.P. is (first term plus last term) times number of terms by 2. In other words \displaystyle { \frac{1}{m_1} + \frac {1}{m_2} + ... + \frac{1}{m_k} = \frac {k}{2} \left ( \frac{1}{m_1} + \frac{1}{m_k} \right)}

So our expression becomes \displaystyle { \frac {k}{ \frac {k}{2} \left ( \frac{1}{m_1} + \frac{1}{m_k} \right)} \le \frac{m_1 + m_2 + ... + m_k}{k} }

This implies \displaystyle { k \le (m_1 + m_2 + ... + m_k) \frac {\left ( \frac{1}{m_1} + \frac{1}{m_k} \right)}{2} }

I cannot think what to do next.

Teacher: It is quite simple actually. Notice that m_1 , ... , m_k are integers then their reciprocals are less than (or at most equal to) 1.

Hence \displaystyle { \frac {1}{m_1} \le 1 , \frac{1}{m_k} \le 1 \Rightarrow \frac{1}{m_1} + \frac{1} {m_k} \le 2 }. m_1 and m_k are different hence both cannot be 1.

Thus \displaystyle { \frac{1}{m_1} + \frac{1} {m_k} < 2 \Rightarrow \frac {\left ( \frac{1}{m_1} + \frac{1}{m_k} \right)}{2} < 1 }

Hence \displaystyle { k \le (m_1 + m_2 + ... + m_k) \frac {\left ( \frac{1}{m_1} + \frac{1}{m_k}\right)}{2} < (m_1 + m_2 + ... + m_k) \le m_1 + k-1} since 1 \le m_2 , ... , 1 \le m_k

Now k is at least 3 (we are already done with k < 3). So k < m_1 + k-1 \le m_1 + 2

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