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I am considering the following sequence with $u_1 > 0 $ and $\alpha > 0$ : $$u_{n+1}=u_n+ \frac{1}{n^{\alpha} u_n}$$

i) I have proven that $(u_n)_n$ converges if and if only $ \alpha >1$. In this case, we write $l$ its limit.

ii) Now I am looking for an asymptotic expansion :

  • for $u_n-l$ if the sequence converges
  • for $u_n$ if the sequence diverges

As $u_{n+1}-u_1=\sum_{k=1}^{n} \frac{1}{ k^{\alpha} u_k}$, I was able to prove the condition of convergence.

iii) I believe that for a sequence $u_{n+1}= f(u_n,n)$, one should find an auxiliary sequence $\lambda_n$ and evaluate $\phi(\lambda_{n+1})-\phi(\lambda_n)$. I consider $v_n=u_n^2$ and try to evaluate $v_{n+1}-v_n$. It looks better but I didn't conclude. More generally, how do I find the auxiliary sequence and $\phi$ ?

Many Thanks,

I made some progress with $v_n=u_n^2$ :

$\begin{align} u_{n+1} &=u_n+ \frac{1}{n^{\alpha} u_n} \\ u_{n+1}^2&=u_n^2 + \frac{1}{n^{2 \alpha} u_n^2} + \frac{2}{n^ {\alpha}} \\ u_{n+1}^2 - u_n^2 &= \frac{1}{ n^{2 \alpha} u_n^2} + \frac{2}{ n ^\alpha } \\ v_{n+1} -v_n &=\frac{1}{ n^{2 \alpha} v_n} + \frac{2}{n^ {\alpha}} \end{align}$

Therefore,

$$ v_{n+1} = \sum_{k=1}^{n} \frac{1}{ k^{2 \alpha} v_k} + \sum_{k=1}^{n} \frac{2}{k^ {\alpha}}$$

zestiria
  • 907
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    have you tried $u_{n+1}-u_n\approx du / dn $? Works usually great with difference equations of this texture... – asgeige Jul 12 '20 at 14:22
  • Taking $u_n = A n^p$ for $0 < \alpha < 1$, $u_n = A n^p \ln^q n$ for $\alpha = 1$ and $u_n = l + A n^p$ for $\alpha > 1$ and applying the dominant balance method gives $$u_n \sim \begin {cases} \sqrt {\frac 2 {1 - \alpha}} , n^{(1 - \alpha)/2} & 0 < \alpha < 1 \ \sqrt {2 \ln n} & \alpha = 1 \ l - \frac 1 {l (\alpha - 1) n^{\alpha - 1}} & \alpha > 1 \end {cases}.$$ – Maxim Jul 17 '20 at 15:49

2 Answers2

1

For $\alpha>1$ we know that $u_n\to L$ so that $u_n=L+o(1)$.

Since we also have

$$u_{n+1}=u_1+\sum_{k=1}^n\frac1{k^\alpha u_k}$$

it follows that

$$L=u_1+\sum_{k=1}^\infty\frac1{k^\alpha u_k}$$

and that the error is given by

\begin{align}L-u_n&=\sum_{k=n}^\infty\frac1{k^\alpha u_k}\\&=\sum_{k=n}^\infty\frac1{k^\alpha(L+o(1))}\\&=\sum_{k=n}^\infty\frac{L^{-1}+o(1)}{k^\alpha}\\&=\frac1{\alpha Ln^{\alpha-1}}+o(n^{1-\alpha})\end{align}

and the asymptotic

$$u_n=L-\frac1{\alpha Ln^{\alpha-1}}+o(n^{1-\alpha})$$

For $\alpha=1$, assume the auxilliary $u_n\sim C[\ln(n)]^\beta$. For $u_{n+1}$ we get

$$C[\ln(n+1)]^\beta=C\left[\ln(n)+\frac1n+\mathcal O(n^{-2})\right]^\beta=C[\ln(n)]^\beta+\frac{\beta C}n[\ln(n)]^{\beta-1}+\mathcal O\left(\frac{[\ln(n)]^{\beta-1}}{n^2}\right)$$

and for $u_n+\frac1{nu_n}$ we get

$$C[\ln(n)]^\beta+\frac1{Cn}[\ln(n)]^{-\beta}$$

Equating coefficients and powers then gives us $\beta=\frac12$ and $C=\frac1{\sqrt2}$, hence

$$u_n\sim\sqrt{\frac{\ln(n)}2}$$

For $0<\alpha<1$, assume assume the auxilliary $u_n\sim Cn^\beta$. For $u_{n+1}$ we get

$$C(n+1)^\beta=Cn^\beta+\beta Cn^{\beta-1}+\mathcal O(n^{\beta-2})$$

For $u_n+\frac1{n^\alpha u_n}$ we get

$$Cn^\beta+\frac1Cn^{-(\alpha+\beta)}$$

Equating coefficients and powers then gives us $\beta=\frac{1-\alpha}2$ and $C=\sqrt{\frac2{1-\alpha}}$, hence

$$u_n\sim\sqrt{\frac{2n^{1-\alpha}}{1-\alpha}}$$

Simply Beautiful Art
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$$u_{n+1}=u_n+ \frac{1}{n^{\alpha}\, u_n} \qquad \text{with} \qquad u_1=a >0$$

If we consider instead (as @asgeige suggested in comments) $$u(n+1)-u(n)= \frac{1}{n^{\alpha}\, u(n)}\sim u'(n)$$ we face a simple differential equation.

If $\alpha=1$ $$\big(u(n)\big)^2=a^2+2 \log (n)$$

If $\alpha>1$ $$\big(u(n)\big)^2=a^2+2\,\frac {1- n^{1-\alpha }}{\alpha-1}$$

Claude Leibovici
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