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Suppose that $\{a_j\}_j$ is a sequence of real numbers. Suppose for all $j$, $a_j \geq 0$ and the sequence $b_j = \frac{a_j}{1 + a_j}$ converges to $0$.

I wish to prove that $a_j$ converges to $0$.

Help would be very much appreciated!

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If $a_j$ does not converge to zero, there is an $\epsilon>0$ such that infinitely often $a_j>\epsilon$. What does this mean for $b_j$? –  Andres Caicedo May 1 '13 at 0:25
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2 Answers

up vote 3 down vote accepted

You don't need the assumption that $a_j \ge 0$.

$$a_j = \frac{1}{1-b_j} - 1$$ for sufficiently large $j$ (why sufficiently large?)

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This is elegant, +1. –  1015 May 1 '13 at 0:39
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Major hint: If $|b_j|<\epsilon$, then $a_j<\epsilon(1+a_j)$. Consequently $a_j(1-\epsilon)<\epsilon$, so $a_j<\frac{\epsilon}{1-\epsilon}$.

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