$u_{n+1}=u_n+ n^{-\alpha} u_n^{-1}$ asymptotic expansion 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,
$$ v_{n+1} = \sum_{k=1}^{n} \frac{1}{ k^{2 \alpha} v_k} + \sum_{k=1}^{n}  \frac{2}{k^ {\alpha}}$$
 A: 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
$$u_n\sim\sqrt{\frac{2n^{1-\alpha}}{1-\alpha}}$$
