As in the title, solve such recurrence: $$\frac{1}{a_{n}}=\frac{1}{a_{n-1}}+\frac{1}{a_{n+1}}$$ for $n\ge 2$, where $a_1=2$ and $a_2=1$.
I mean, any hints?
Solving a recurrence: $\frac{1}{a_{n}}=\frac{1}{a_{n-1}}+\frac{1}{a_{n+1}}$
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recurrence-relations
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3Set $b_n = \frac1{a_n}$ to reduce the problem to its essence. – 2017-01-19
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0Guess it's too late for me now, this was really obvious. – 2017-01-19
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1 Answers
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Hint: Can you solve $$b_{n+1} = b_n - b_{n-1}\qquad n\geq 2$$ where $b_1=\frac{1}{2}$, $b_2=1$?
If so, can you define $(b_n)_n$ with regard to $(a_n)_n$ so that solving the above is equivalent to solving your original problem?
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0Guess it's too late for me now, this was really obvious. – 2017-01-19
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2@VanDerWarden To be honest, given the "right" amount of sleep-deprivation nothing is ever completely obvious. – 2017-01-19