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The sequence converges to zero at a rate that seems to be slightly faster than $1/n$. What are the best known results on the convergence rate of this sequence?

yoyoyoe
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3 Answers3

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You can use Cesaro's theorem: Put $\displaystyle u_n=\frac{1}{a_{n+1}}-\frac{1}{a_n}$. We compute easily that $\displaystyle u_n=\frac{1}{1-a_n}$, hence $u_n\to 1$. By Cesaro's theorem, we get that $\displaystyle \frac{u_1+\cdots+u_n}{n}\to 1$. This gives with easy computation that $n a_n\to 1$, hence $\displaystyle a_n\sim \frac{1}{n}$.

Kelenner
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  • Thanks Kelenner, this is indeed helpful! This shows that rate is $\frac{1}{n+o(n)}$, but maybe there is a way to tell something more precise about the $o(n)$ part? – yoyoyoe Dec 03 '15 at 20:44
  • I have tried, but sorry, I do not see how to say something on the $o(n)$ part. Maybe the following formula is interesting: as $a_{n+1}-a_n=-a_n^2$, you get $a_{n+1}=a_1-\sum_{k=1}^{n}a_k^2$, and as $a_{n+1}\to 0$, $\sum_{k\geq 1} a_k^2=a_1$, hence $a_{n+1}=\sum_{k\geq n+1} a_k^2$. – Kelenner Dec 03 '15 at 21:11
  • Very impressive approach! –  Dec 03 '15 at 21:15
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The sequence is a monotonic decreasing one. Also, it is easy to see that it is always positive. Hence, $\{a_n\}$ converges.

Let $N$ be sufficiently large integer so that $$ a_N - a_{N + 1} < \varepsilon_N, $$ where $\varepsilon_n \to 0$. But we have $$ a_N - a_{N + 1} = a_N (1 - 1 + a_N) = a_N^2 < \varepsilon_N. $$ Hence, $a_N \to 0$ by squeeze theorem.

Now let $r_n = 2^{2^n}$, $b_n = b_{n - 1} (r_{n - 1} - b_{n - 1})$ and $b_0 = 1$. Then, $$ a_n = \frac {b_n} {r_n}. $$ Since $$ \left(\frac {r_{k - 1}} {2} - b_{k - 1}\right)^2 = \frac {r_k} {4} - b_k $$ for every nonnegative integer $k$, we deduce that $$ \sum_{k = 1}^{n} \left(\frac {r_{k - 1}} {2} - b_{k - 1}\right)^2 = \sum_{k = 1}^{n} \frac {r_k} {4} - \sum_{k = 1}^{n} b_k, $$ and, by Cauchy-Schwartz inequality, we have $$ \sum_{k = 1}^{n} \frac {r_k} {4} - \sum_{k = 1}^{n} b_k > \frac {1} {n} \left (\sum_{k = 1}^{n} \frac {r_{k - 1}} {2} - \sum_{k = 1}^{n} b_{k - 1}\right)^2. \qquad \qquad (1) $$ Denote $$ R_n = r_1 + r_2 + \cdots + r_{n - 1}, \qquad B_n = b_1 + b_2 + \cdots + b_{n - 1}, \nonumber \\ \phi (n) = \sqrt {n \left (n + r_n - R_n + \frac {3} {2} - 4 b_n\right) - R_n}. $$ After simplification, $(1)$ becomes $$ \frac {2 R_n - 2n - 3} {4} - \frac {\phi (n)} {2} < B_n < \frac {2 R_n - 2n - 3} {4} + \frac {\phi (n)} {2}. $$ Taking into account that $R_n = o (r_n)$ and $b_n = B_{n + 1} - B_n$, we have $$ \frac {r_n} {n} < b_n < \frac {r_n} {n} (1 + o (1)). $$ Hence, $a_n = \frac {1} {n} + o \left(\frac {1} {n}\right)$.

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Here is a slightly better bound: first, it is more convenient to look at $b_n = 1/a_n$. We have $$\frac{1}{b_{n+1}}=\frac{1}{{b_n}}-\frac{1}{{b_n^2}}=\frac{b_n-1}{b_n^2}$$ or, $$b_{n+1}=\frac{b_n^2}{b_n-1}=\frac{(b_n+1)(b_n-1)+1}{b_n-1}=b_n+1+\frac{1}{b_n-1}.$$ It is easy to see that $b_k\geq 2$ for all $k$, which implies $$b_{n}\leq b_{n-1}+1+\frac{1}{2-1}\leq\dots\leq 2n+2.$$ Now, $$ b_{n+1}\geq b_n+1+\frac{1}{1+2n}\geq b_{n-1}+2+\frac{1}{2n+1}+\frac{1}{2n-1}\geq \dots\geq b_0+n+1+\sum_{k=0}^n \frac{1}{2k+1}, $$ and since $$ \sum_{k=0}^n \frac{1}{2k+1} > \frac{1}{2}\sum_{k=0}^n \frac{1}{k+1} \geq \frac{1}{2}\ln(n+2) $$ we get $$ b_{n+1}\geq n+\frac{1}{2}\ln(n+2)+3. $$ Using the same reasoning, from $b_n\geq n+2$ (and using $\sum_{k=1}^n \frac{1}{k} \leq \ln n+1$) we get $$ b_{n+1}\leq n+\ln(n+1)+4, $$ hence $$ \frac{1}{n+\ln(n+1)+4}\leq a_{n+1} \leq \frac{1}{n+\frac{1}{2}\ln(n+2)+3}. $$

yoyoyoe
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