asymptotics
Below is a question I faced from an online test for preparation of exam and I had doubt in solution provided so I wanted to discuss my approach and ask about it.
$$\frac{e^{n\log n}}n(A),n^{\sqrt n}(B),2^{n\log n}(C)$$
I want to order these functions based on increasing asymptotic order.
I rewrite A as $\frac{n^{nlog e}}{n}=\frac{n^{1.44n}}{n}=n^{1.44n-1}$
C is re-written as $n^{nlog2}=n^n$
So, finally, my order came to be A, B, C in increasing asymptotic growth.
Am I correct?
I'd rather write $$2^{\sqrt{\log_2n}}/\sqrt n=2^{\sqrt{\log_2n}-(1/2)\log_2n}$$ and then note $$\sqrt{\log_2n}-(1/2)\log_2n\to-\infty$$ as $n\to\infty$ so the limit is, indeed, zero. This doesn't prove $2^{\sqrt{\log_2n}}\lt\sqrt n$ for all $n$, but certainly implies that inequality for all sufficiently large $n$.
Define $P(x)=c_2 x^2+c_1 x+c_0$.
Rewrite the problem as $$w(\eta)\sim\frac1{\Gamma(\alpha)}\int^\eta_0 ds\frac{(1-s/\eta)^{\alpha-1}}{(s+\eta_0)^{\alpha+1}}P\circ w(s)$$
As a shorthand, define the operator $$I=\int^\eta_0 ds\frac{(1-s/\eta)^{\alpha-1}}{(s+\eta_0)^{\alpha+1}}$$
Also define $A$ as the 'asymptotic operator', such that $A[f]$ returns the asymptotics of $f$ up to $o(1)$, according to certain asymptotic scale.
Then, the problem becomes a fixed-point problem: $$w=\frac1{\Gamma(\alpha)}\cdot A\circ I\circ P\circ w\qquad(\star)$$
It is assumed that $w(\eta)$ grows at most polynomially as $\eta\to\infty$.
First, we investigate $A\circ I$'s action on a monomial $x^q$ ($q\ge0$).
$$\begin{align} I[s^q](\eta) &=\int^\eta_0\frac{(1-s/\eta)^{\alpha-1}}{(s+\eta_0)^{\alpha+1}}s^q ds \\ &=\int^\eta_0\frac{s^q}{(s+\eta_0)^{\alpha+1}}ds+ \underbrace{\int^\eta_0\frac{(1-s/\eta)^{\alpha-1}-1}{(s+\eta_0)^{\alpha+1}}s^q ds}_{I'(q)} \\ \end{align} $$
For $I'(q)$, $$\begin{align} I'(q) &=\int^{\eta}_0 \frac{(1-s/\eta)^{\alpha-1}-1}{(s+\eta_0)^{\alpha+1}}s^q ds \\ &=\sum^\infty_{r=1}\binom{\alpha-1}{r}(-1)^r \eta^{-r}\int^\eta_0\frac{s^{q+r}}{(s+\eta_0)^{\alpha+1}}ds \\ &=\sum^\infty_{r=1}\binom{\alpha-1}{r}(-1)^r \eta^{-r}\left(\frac{\eta^{q+r-\alpha}}{q+r-\alpha}+O(\eta^{q+r-\alpha-1})\right) \\ &=\eta^{q-\alpha}\sum^\infty_{r=1}\binom{\alpha-1}{r}\frac{(-1)^r}{q+r-\alpha}+O(\eta^{q-\alpha-1}) \\ &=\eta^{q-\alpha}\left[\frac{\Gamma(\alpha)\Gamma(q-\alpha)}{\Gamma(q)}-\frac1{q-\alpha}\right]+O(\eta^{q-\alpha-1}) \\ \end{align} $$
Note that $\gamma(q):=\frac{\Gamma(\alpha)\Gamma(q-\alpha)}{\Gamma(q)}-\frac1{q-\alpha}$ converges for $\Re (q)\ge 0$.
Summary $$I[s^q](\eta)=\int^\eta_0\frac{s^q}{(s+\eta_0)^{\alpha+1}}ds+\gamma(q)\eta^{q-\alpha}+O(\eta^{q-\alpha-1})$$
By differentiating under the integral sign, it can be further proved that
$$I[s^q \ln^N s](\eta)=
\begin{cases}
O(\eta^{q-\alpha}\ln^N \eta), &q>\alpha \\
O\left(\ln^{N+1}\eta\right), &q=\alpha \\
O(1), &\alpha> q\ge 0
\end{cases}
$$
(Asymptotic expansions up to $o(1)$ are given in the appendix below.)
It is obvious that, to match the leading order on both sides of $(\star)$, $w(\eta)\sim C\eta^{\alpha}$ where $C=\frac{\Gamma(2\alpha)}{c_2 \Gamma(\alpha)}$, as the OP obtained.
Now, let's make a slightly different ansatz: $$w(\eta)=C\eta^{\alpha}+\sum^N_{r=0}B_r\ln^r\eta$$
Utilizing asymptotic formulae in the appendix, ($D:=\frac{2\Gamma(2\alpha)}{\Gamma^2(\alpha)}$) $$\begin{align} \frac1{\Gamma(\alpha)}\cdot A\circ I\circ P\circ w &=C\eta^{\alpha}+\frac{DB_N}{N+1}\ln^{N+1}\eta \\ &+D\sum^N_{r=2}\left[\frac{B_{r-1}}{r}+\sum^N_{m=r}\binom{m}{r}\gamma^{(m-r)}(\alpha)B_m\right]\ln^r\eta \\ &+D\left[\frac{c_1\Gamma(\alpha)}{2c_2}+2B_0+\sum^N_{m=1}m\gamma^{(m-r)}(\alpha)B_m\right]\ln\eta \\ &+\text{a messy bunch of constants} \end{align} $$ (The messy bunch of constants can be found using the asymptotic formulae in appendix.)
The mismatch of order, i.e. the presence of $\ln^{N+1}\eta$, is expected. However, in the following calculations, we will give evidence hinting at rapid decay of $B_r$, such that
If these two conditions are satisfied, an ansatz with $N=\infty$ is a solution to $(\star)$.
Claim $$ B_{N-i}=c_i\cdot\frac{N!}{(N-i)!}B_N \qquad (\blacksquare)$$ where $c_i$ is independent of $N$.
By comparing coefficients, $$B_r=\frac{B_{r-1}}{r}+\sum^N_{m=r}\binom{m}{r}\gamma^{m-r}(\alpha)B_m$$ $$\implies \frac{B_{r-1}}{r}+(\gamma(\alpha)-1)B_r+\sum^N_{m=r+1}\binom{m}{r}\gamma^{m-r}(\alpha)B_m=0 \qquad (1)$$
Let $r=N-i$. After a re-indexing of the sum, $$\frac{B_{N-i-1}}{N-i}+(\gamma(\alpha)-1)B_{N-i}+\sum^{i-1}_{m=0}\binom{N-m}{N-i}\gamma^{i-m}(\alpha)B_{N-m}=0$$
Substitute in $B_{N-i}=c_i\cdot\frac{N!}{(N-i)!}B_N$,
$$\begin{align} &c_{i+1}\frac{N! B_N}{(N-i)!}+(\gamma(\alpha)-1)c_i\frac{N! B_N}{(N-i)!} \\ &+\sum^{i-1}_{m=0}\frac{(N-m)!}{(N-i)!(i-m)!}\cdot \gamma^{i-m}(\alpha)\cdot c_m\frac{N!B_N}{(N-m)!} = 0 \end{align} $$
Here we notice mass cancellations, leading to $$c_{i+1}=(1-\gamma(\alpha))c_i-\sum^{i-1}_{m=0}\frac{\gamma^{(i-m)}(\alpha)}{(i-m)!}c_m$$
Absence of $N$ supports our claim.
Furthermore, by a contour integration approach, it can be proved that $$\frac{\gamma^{(x)}(\alpha)}{x!}=\frac{(-1)^{x}}{\Gamma(1-\alpha)}\sum^{\infty}_{n=1}\frac{\Gamma(n+\alpha+1)}{n! n^{x+1}}$$ for $x\ge 1$.
This can be used to estimate $c_i\approx R^i$ for some constant $R>1$.
Thus, we conclude that:
Practically, most Volterra integral equations are solved numerically. Therefore, it is feasible to directly plug in the infinte series and solve for the coefficients numerically. Noteworthily, viewing $B_r$ as unknowns, $(1)$ creates a system of linear equations, with an almost triangular coefficient matrix and the constant matrix $\begin{pmatrix} -B_0 & 0 & \cdots & 0 \end{pmatrix}^T$. Solving the system seems like an easy task for computer.
(This explains why $B_0$ (or $\lambda$) cannot be zero: otherwise the system of equations becomes homogeneous, and since the coefficient matrix is invertible, all $B_r=0$. The solution becomes trivial.)
A very special solution exists when $\displaystyle{\eta_0=\left[\frac{c_1^2(1-c_2)-c_0}{c_1\Gamma(\alpha+1)}\right]^{1/\alpha}}$: $$w(\eta)=C\eta^{\alpha}-c_1$$
One approach that I haven't tried is substituting $\eta=e^z$ - the infinte series then becomes a Maclaurin series, whose coefficients may be found (probably) by Cauchy's integral formula or simply by differentiation. I may add on this later.
Define $v(z)=w(e^z)$.
Substitute in $\eta=e^z$, $$v(z)=\frac{e^z}{\Gamma(\alpha)}\int^1_0 \frac{(1-s)^{\alpha-1}}{(e^z s+\eta_0)^{\alpha+1}}P\circ v(z+\ln s)ds$$
The ansatz is $$v(z)=\sum^\infty_{i=0}a_i z^i$$
Let's start from the right hand side.
$$\begin{align} v(z+\ln s) &=\sum^\infty_{i=0}\sum^i_{j=0}a_i\binom{i}{j}(\ln s)^{i-j}z^j \\ &=\sum^\infty_{j=0}\left[\sum^\infty_{i=j}a_i\binom{i}{j}(\ln s)^{i-j}\right]z^j \\ &=\sum^\infty_{j=0}\sum^\infty_{i=0}a_{i+j}\binom{i+j}{j}(\ln s)^{i}z^j \\ &=\sum^\infty_{j=0}\sum^\infty_{i=0}A[i,j](\ln s)^{i}z^j \\ \end{align} $$ where $A[x,y]=a_{x+y}\binom{x+y}{y}$.
Viewing this double series as a 'two-dimensional' Taylor series, Cauchy product immediately gives $$v^2(z+\ln s)=\sum^\infty_{i=0}\sum^\infty_{j=0}(A*A)[i,j](\ln s)^{i}z^j \\$$ where $(A*A)[x,y]=\sum^x_{m=0}\sum^y_{n=0}A[m,n]A[x-m,y-n]$ is the discrete convolution.
Hence, $$P\circ v(z+\ln s)=c_0+\sum^\infty_{i=0}\sum^\infty_{j=0}\underbrace{\bigg(c_2(A*A)[i,j]+c_1A[i,j]\bigg)}_{B_{ij}}(\ln s)^i z^j$$
Operating $\frac1{\Gamma(\alpha)}\cdot A\circ I$ on $P\circ v(z+\ln s)$, we can finally convert $(\star)$ into
$$\Gamma(\alpha)\sum^\infty_{j=0}a_j z^j =\frac{c_0\eta_0^{-\alpha}}{\alpha} +\sum^\infty_{j=0}\sum^\infty_{i=0}B_{ij}\sum^i_{k=0}\binom{i}{k}\int^\infty_0\frac{\ln^{i-k}s}{(s+\eta_0)^{\alpha+1}}ds \cdot z^{j+k}$$
which is suitable for coefficient comparison and solving by computer. At last, $w(\eta)=v(\ln\eta)$.
For $\alpha<q\le 2\alpha$,
$$I[s^q \ln^N s](\eta)=\int^\eta_0\frac{s^q\ln^N s}{(s+\eta_0)^{\alpha+1}}ds+\eta^{q-\alpha}\sum^N_{r=0}\binom{N}{r}\gamma^{(m-r)}(q)\ln^r\eta+O(\eta^{q-\alpha-1}\ln^N \eta)$$
For $q=\alpha$,
$$\begin{align} I[s^\alpha \ln^N s](\eta) &=\frac{\ln^{N+1}\eta}{N+1}+\eta_0\cdot\frac{\alpha+1}{N+1}\int^\infty_0\frac{s^\alpha\ln^{N+1}s}{(s+\eta_0)^{\alpha+2}}ds \\ &+\sum^N_{r=0}\binom{N}{r}\gamma^{(m-r)}(\alpha)\ln^r\eta+O(\eta^{-1}\ln^N \eta) \end{align} $$
For $0\le q<\alpha$,
$$I[s^q \ln^N s](\eta)=\int^\infty_0\frac{s^q\ln^N s}{(s+\eta_0)^{\alpha+1}}ds+O(\eta^{q-\alpha}\ln^N\eta)$$
Best Answer
$$\frac{e^{n\log_2n}}n=\frac{e^{n(\ln n)/(\ln 2)}}n=\frac{n^{n/\ln 2}}n=n^{n/\ln 2-1}$$ $$2^{n\log_2 n}=n^n$$ We now compare powers: $\sqrt n<\frac n{\ln 2}-1<n$. Thus $B<C<A$, not $A<B<C$ as you claimed.