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Let $\{a_n\}_{n=0}^\infty :$ $$a_0 = 1, a_1 = \frac{1}2, a_{n+1} = \frac{n a_n^2}{1 + (n+1) a_n}, n \ge 1$$

How to prove that series $\sum_{n=0}^\infty \frac{a_{n+1}}{a_n}$ is convergent and calculate its sum?

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$$ a_{n+1} + (n+1) a_n a_{n+1} - n a_n^2 = 0 \tag{1}$$ $$ \frac{a_{n+1}}{a_n} = n a_n - (n+1) a_{n+1}\tag{2} $$ hence it is enough to prove that $n a_n \to 0$ to get from $$ \sum_{n=0}^{N}\frac{a_{n+1}}{a_n} = 1-(N+1)a_{N+1}\tag{3} $$ that the value of the given series is simply $1$.

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    $\begingroup$ Elegant proof! (+1) And for $b_n := n a_n \to 0$, we can for instance use the inequality $$ b_{n+1} = \frac{b_n^2}{\frac{n}{n+1} + b_n} \leq \frac {b_n^2}{2\sqrt{\frac{n}{n+1}b_n}} \leq b_n^{3/2}, \quad b_1 = \tfrac{1}{2} $$ to get a very rapid convergence. $\endgroup$ – Sangchul Lee Nov 10 '16 at 20:35
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To complement Jack D'Aurizio's excellent observation: define $b_n = na_n$. Since $$ b_{n+1} = \frac{(1+\frac1n)b_n^2}{1+(1+\frac1n)b_n} < b_n, $$ we see that $\{b_n\}$ is a decreasing sequence that is bounded below by $0$. Therefore $\{b_n\}$ converges to some limit, say $L$. Then $$ L = \lim_{n\to\infty} b_{n+1} = \lim_{n\to\infty} \frac{(1+\frac1n)b_n^2}{1+(1+\frac1n)b_n} = \frac{(1+0)L^2}{1+(1+0)L} = \frac{L^2}{1+L}, $$ and the only solution is $L=0$.

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