Prove that $ t_{n}=1+\frac{\ln (2 n)}{2 n}+o\left(\frac{1}{n}\right) $ as $n$ tends to infinity. For all $n \geqslant 1$, we define
$$
f_{n}(t):=t^{2 n}-2 n t+1
$$
(i) Prove that there exists a unique solution of $f_{n}(t)=0$ in $[1,+\infty)$. This solution will be denoted by $t_{n}$.
(ii) Prove that $\lim _{n \rightarrow+\infty} t_{n}=1$.
(iii) Prove that
$$
t_{n}=1+\frac{\ln (2 n)}{2 n}+o\left(\frac{1}{n}\right)
$$
as $n$ tends to infinity.

For the question (I)
Since $\lim_{t\to 1^{+}} f_n(t)=-2n<0$ And $\lim_{t\to +\infty}f_n(t)=+\infty<0$
By intermediate value theorem there exist $t_n \in[1, +\infty)$ which $f_n(t_n)=0$
For other question, I think for a lot but I can’t do it. please kindly give me a hint or something that I can do this. THANK in advance !
 A: For (iii) (which proves (ii) as well): First note that
$$
t_n^{2n}  + 1 = 2nt_n  \ge 2n,
$$
i.e.,
$$
t_n  \ge \sqrt[{2n}]{{2n - 1}}.
$$
But
$$
\sqrt[{2n}]{{2n - 1}} = \exp \left( {\frac{\log (2n - 1)}{{2n}}} \right) = 1 + \frac{\log (2n - 1)}{{2n}} + o\!\left( {\frac{1}{n}} \right) = 1 + \frac{\log (2n)}{{2n}} + o\!\left( {\frac{1}{n}} \right).
$$
Thus,
$$\tag{1}
t_n  \ge 1 + \frac{\log (2n)}{{2n}} + o\!\left( {\frac{1}{n}} \right).
$$
We also have
$$
2nt_n  = t_n^{2n}  + 1 \ge t_n^{2n} ,
$$
i.e.,
$$
\sqrt[{2n - 1}]{{2n}} \ge t_n .
$$
But
$$
\sqrt[{2n - 1}]{{2n}} = \exp \left( {\frac{{\log (2n)}}{{2n - 1}}} \right) = 1 + \frac{{\log (2n)}}{{2n - 1}} + o\!\left( {\frac{1}{n}} \right) = 1 + \frac{{\log (2n)}}{{2n}} + o\!\left( {\frac{1}{n}} \right).
$$
Hence,
$$\tag{2}
t_n  \le 1 + \frac{\log (2n)}{{2n}} + o\!\left( {\frac{1}{n}} \right).
$$
From $(1)$ and $(2)$,
$$ 
t_n = 1 + \frac{\log (2n)}{{2n}} + o\!\left( {\frac{1}{n}} \right),
$$ as desired.
A: This question bears a striking resemblance to a problem I recently posted on the arXiv. At the risk of sounding self promoting, I encourage you to check out the paper as many of the techniques for solving this problem can be found in the paper in great detail.
As for the answer to your question, first consider the special case $n=1$, for which we can use the quadratic formula to show $f_n(t)$ has exactly one root at $t_1=1$. Now consider the remaining case $n>1$ and note that the signs of the coefficients of $f_n(t)=t^{2n}-2nt+1$ in order of descending variable exponent gives the sequence $(+1,-1,+1)$ showing two variations in sign. By Descartes' rule of signs $f_n(t)=0$ has exactly two or zero solutions on the positive reals $t\in[0,\infty)$. But $f_n(0)=1>0$, $f_n(1)=2(1-n)<0$, and $\lim_{t\to\infty}f_n(t)=\infty>0$; thus, we have one root on the interval $t\in[0,1)$ and another root on $t\in[1,\infty)$ for all $n>1$. This proves part $(i)$ of your problem.
I now will provide an outline of how to solve the remaining two parts based on the process outlined in the paper.  Using Lagrange inversion we may find an explicit series representation for $t_n$ (this is left as an exercise). To obtain the limiting value of $t_n$ and its asymptotic expansion we first use a little algebra to show that $t_n$ satisfies
$$
\frac{1}{2n}t_n^{2n}+(t_n-1)-(1-\tfrac{1}{2n})=0.
$$
The ansatz $t_n\to 1$ as $n\to\infty$ then suggests we study the perturbed problem
$$
\tag{1}
\frac{1}{2n}t_n^{2n}+(t_n-1)\epsilon-(1-\tfrac{1}{2n})=0
$$
and derive a perturbation series expansion for $t_n$ in $\epsilon$ of the form $t_n=\sum_{k=0}^\infty a_k\epsilon^k$. Evaluating the perturbation series at $\epsilon=1$ then recovers the exact value of $t_n$. Using the well known formula for integer powers of power series we have
$$
t_n^{2n}=\sum_{k=0}^\infty c_k\epsilon^k,
$$
where $c_0=a_0^{2n}$ and $c_k$ satisfies a recurrence relation (see paper for recurrence relation). Substituting the perturbation series for $t_n$ into $(1)$ and collecting like powers of $\epsilon$ we find
$$
\tfrac{1}{2n} a_0^{2n}-(1-\tfrac{1}{2n})+(c_1+a_0-1)\epsilon+\sum_{k=1}^\infty(c_{k+1}+a_k)\epsilon^{k+1}=0,
$$
Equating the coefficients of $\epsilon^k$ to zero yields an infinite system of equations for determining the coefficients $a_k$. For example, setting the constant term equal with zero we find
$$
\tfrac{1}{2n} a_0^{2n}-(1-\tfrac{1}{2n})\implies a_0=(2n-1)^{\frac{1}{2n}}.
$$
Now, I will not find the rest of the $a_k$ here but you can follow the process outlined in the arXiv paper to obtain an exact result for them.
What I claim is that for all $k\geq 1$ we have $\lim_{n\to\infty}a_k=0$ so that $t_n\sim a_0$ as $n\to\infty$. Indeed we find
$$
\lim_{n\to\infty}a_0=1
$$
and as $n\to\infty$
$$
a_0\sim 1+\frac{\log(2n)}{2n}+o\left(\frac{1}{n}\right),
$$
which is the desired conclusion for parts $(ii)$ and $(iii)$.
A: For part two, just play around a little bit:
$$2nt = t^{2n}+1 \geq 2t^n\implies t^{n-1}\leq n\implies ...$$
For part three, you need to play around a little bit more carefully. Writing $t = 1 + x$ and then Taylor expanding around $x = 0$, should give you some ideas.
A: As $f$ is increasing on $[1,\infty)$, we can prove that for $n>1$:
$$1+\frac{\ln(2n)}{2n}<t_n<1+\frac{\ln(2n)}{2n-2}$$
using that for $x>0$, $\left(1+\frac{x}{n}\right)^n\uparrow e$ and $\left(1+\frac{x}{n-1}\right)^n\downarrow e$ as $n\to\infty$.
Indeed:
$$\left(1+\frac{\ln(2n)}{2n}\right)^{2n}-2n\left(1+\frac{\ln(2n)}{2n}\right)+1<e^{\ln(2n)}-2n\left(1+\frac{\ln(2n)}{2n}\right)+1=1-\ln(2n)<0$$
and:
$$\left(1+\frac{\ln(2n)}{2n-2}\right)^{2n}-2n\left(1+\frac{\ln(2n)}{2n-2}\right)+1=\left(1+\frac{\ln(2n)}{2n-2}\right)^{2n-1}\left(1+\frac{\ln(2n)}{2n-2}\right)-2n\left(1+\frac{\ln(2n)}{2n-2}\right)+1>e^{\ln(2n)}\left(1+\frac{\ln(2n)}{2n-2}\right)-2n\left(1+\frac{\ln(2n)}{2n-2}\right)+1=1>0$$
Therefore:
$$0<t_n-1-\frac{\ln(2n)}{2n}<\frac{\ln(2n)}{2n(n-1)}$$
From which (ii) and (iii) follow easily.
