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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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