Limit of a sequence satisfying a given equation Find the limit of the sequence $(x_n)\subset (0,1)$ such that
$$
\frac{1-x_n^n}{n}-\frac{1-x_n^{n+1}}{n+1}=\frac{1}{2n(n+1)}.
$$
I tried to sum the equation for $n$ from $1$ to $p>1$, and my hope was to get a telescoping series, but I didn't get anything like this.
Also I proved that the sequence $(x_n)$ is monotonical increasing and my guess it that its limit is $1$. Am I missing something? How can I overpass the restriction given by the equation above?
 A: Suppose that $x_n$ satisfies $\frac{1-x_n^n}{n} - \frac{1-x_{n}^{n+1}}{n+1} = \frac 1{2n(n+1)}$. This rearranges upon cross-multiplication to $$
1- (n+1)x_n^n + nx_n^{n+1}=\frac 12 \implies (n+1)x_n^n-nx_n^{n+1} = \frac 12 \implies x_n^n((n+1)-nx_n) = \frac 12
$$
Now, suppose that $x_n$ has a convergent subsequence to say $L$. Then, we roughly have $x_n^n \approx L^n$ and$(n+1) -nx_n \approx (n+1)-nL \approx n(1-L) - 1$.
Substituting this in the equation tells us that $$
L^n(n(1-L)-1) \approx \frac 12
$$
should hold for large enough $n$. As the $x_n \in (0,1)$ by choice, we can only have $L \in [0,1]$. Observe that there are two terms, $L^n$ and $n(1-L)-1$ in this description. The product of these two is to remain constant as $n$ increases.
However, this is impossible if $L<1$. Indeed, if $L<1$ then the term $L^n$ decreases exponentially, by a factor of $L$ each time. On the other hand, the term $n(1-L)-1$ increases, but only linearly because there is a constant $1-L$ addition at each term. Thus, their product being held constant is untenable (a logic similar to the limit $\lim_{n \to \infty} na^n=0$ for $a \in (0,1)$ gives this). It follows that $L=1$ is the only possibility.
Now, we just have to make this rigorous carefully.

We will use the following fact :

Let $x_n$ be a sequence such that, for some fixed finite $L$, every subsequence $x_{n_k}$ of $x_{n}$ has a further subsequence $x_{n_{k_m}}$ which converges to $L$. Then, $x_n$ itself converges to $L$.

So we start with a subsequence $x_{n_k}$ of $x_n$. As it is bounded, by the Bolzano-Weierstrass theorem it has a convergent subsequence say $x_{n_{k_l}}$. For convenience, let $m_l = n_{k_l}$. We will prove that $x_{m_l} \to 1$ so that $L=1$ works.
To prove that $x_{m_l} \to 1$, we will make the "growth" argument rigorous by exactly seeing how much we lose by the "replacement" of $x_n$ with $L$ in the previous section.
Assume without loss of generality that $L<1$ is the limit of the $x_{m_l}$. We write $\epsilon_l = x_{m_l}-L$. Then, $$
\frac 12 = x_{m_l}^{m_l}((m_l+1 - m_lx_{m_l})) = (L+\epsilon_l)^{m_l}(m_l(1-L-\epsilon_l)+1)
$$
We play a common binomial trick at this point. Let $L+\epsilon_l = \frac 1{1+t_l}$ where $t_l>1$ (note that $\epsilon_l \to 0$ so any large enough $l$ will have an associated $t_l$ when $L+\epsilon_l < 1$). You can easily find $t_l$ in terms of $\epsilon_l$ and $L$, it's a quantity such that $t_l \to \frac{1}{L}-1$ as $l \to \infty$.
Then, by taking merely one term of the binomial theorem in the expansion of $(1+t_l)^{m_l}$, we have $$
(1+t_l)^{m_l} > \frac{m_l(m_l-1)}{2}t_l^2 \implies (L+\epsilon_l)^{m_l} < \frac{2}{m_l(m_l-1)t_l^2}
$$
Hence, $$
\frac 12 = (L+\epsilon_l)^{m_l}(m_l(1-L-\epsilon_l)+1) < \frac{2}{m_l(m_l-1)t_l^2}(m_l(1-L-\epsilon_l)+1)\\ < \frac{2(1-L-\epsilon_l)}{(m_l-1)t_l^2} + \frac{2}{m_l(m_l-1)t_l^2}
$$
However, the right hand side converges to $0$ as $l \to \infty$. It follows that $L<1$ is impossible.
However, $x_{m_l}$ is a sequence that is bounded between $0$ and $1$. The limit points of $(0,1)$ all lie in the set $[0,1]$. It follows that $x_{m_l} \to 1$. As the subsequence $x_{n_k}$ was arbitrary, applying the fact gives the theorem.
The proof of the fact is a standard exercise : if $x_n$ doesn't converge to $L$ then for some $\epsilon>0$ , there is no $N$ such that $n>N$ implies that $|x_n - L| > \epsilon$. Start with any $n_1 = N$ such that $|x_N - L| \geq \epsilon$. Then find a larger $n_2>n_1$ such that the same things holds and keep going, which you can because you assumed that there is no $N$ beyond which the opposite of the inequality holds. That subsequence $n_1,n_2,n_3,\ldots$ cannot have a subsequence converging to $L$, a contradiction to our assumption. Hence the fact is proved.
