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I am triyng to solve this interesting problem, but I have no idea. I tried induction, but it didn't work. Let $(x_n)_{n \ge 1}$ be a sequence, where $x_1=1$ and defined as follows: $$x_{n+1}=x_n+ \frac {1} {3x_n^2},$$ for every $n \ge 1$. Show that there exists a real number $c \gt 0$ so as $x_n \lt \sqrt[3] {n+c}.$

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The statement is wrong. Define $y_n = x_n^3$. Then $y_1 = 1$ and $$ y_{n+1} = y_n + 1 + \frac{1}{3 y_n} + \frac{1}{27 y_n^2} > y_n + 1 + \frac{1}{3 y_n} \, . $$ It follows that $$ y_n = y_1 + \sum_{k=1}^{n-1}(y_{k+1} - y_k) > 1 + \sum_{k=1}^{n-1}\left(1 + \frac{1}{3 y_k}\right) = n + \frac 13 \sum_{k=1}^{n-1} \frac{1}{y_k} \\ \Longrightarrow y_n - n > \frac 13 \sum_{k=1}^{n-1} \frac{1}{y_k} \, . $$

Now assume that $y_n < n + c$ for some real number $c > 0$ and all $n \in \Bbb N$. Then $$ c > y_n - n > \frac 13 \sum_{k=1}^{n-1} \frac{1}{y_k} > \frac 13 \sum_{k=1}^{n-1} \frac{1}{k + c} $$ which is a contradiction because the series on the right-hand side diverges for $n \to \infty$.

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