# Find $\sum_{k=1}^\infty\frac{1}{x_k^2-1}$ where $x_1=2$ and $x_{n+1}=\frac{x_n+1+\sqrt{x_n^2+2x_n+5}}{2}$ for $n \ge 2$

Given $$x_1=2$$ and $$x_{n+1}=\frac{x_n+1+\sqrt{x_n^2+2x_n+5}}{2}, n\geq 2$$

Prove that $$y_n=\sum_{k=1}^{n}\frac{1}{x_k^2-1}, n\geq 1$$ converges and find its limit.

1. To prove a convergence we can just estimate $$x_n > n$$, therefore $$y_n, where $$z_n=\sum_{k=1}^{n}\frac{1}{k^2-1}$$ and $$z_n$$ converges, then $$y_n$$ converges too.

2. We can notice that $$x_n^2+2x_n+5=(x_n+1)^2+4$$. So $$x_{n+1}$$ is one of the roots of the equation: $$x_{n+1}^2-(x_n+1)x_{n+1}-1=0$$
So $$x_{n+1}^2-1=(x_n+1)x_{n+1}$$ and therefore: $$y_n=\sum_{k=1}^n \frac{1}{(x_{n-1}+1)x_{n}}$$

I'm stuck here.

• The first term in your $z_n$ is undefined.
– Gary
May 22 at 0:01

Using the relation $$x_{n+1}^2 - 1 = x_{n+1}(x_n + 1)$$, we find that

\begin{align*} \frac{1}{x_n + 1} - \frac{1}{x_{n+1} + 1} &= \frac{x_{n+1}}{x_{n+1}^2 - 1} - \frac{1}{x_{n+1} + 1} \\ &= \frac{1}{x_{n+1}^2 - 1}. \end{align*}

So it follows that

\begin{align*} y_n &= \frac{1}{x_1^2 - 1} + \sum_{k=1}^{n-1} \left( \frac{1}{x_k + 1} - \frac{1}{x_{k+1} + 1} \right) \\ &= \frac{1}{x_1^2 - 1} + \frac{1}{x_1 + 1} - \frac{1}{x_n + 1} \\ &\xrightarrow[n\to\infty]{} \frac{x_1}{x_1^2 - 1} = \frac{2}{3}. \end{align*}

• Great solution! But you put a "-" instead of a "+" in $\frac{1}{x_{k+1}+1}$ term. May 21 at 23:45
• @pelfox, Thank you! I fixed my answer accordingly. May 21 at 23:49

Making use of the following equality, and standard partial fraction decompositions: $$x_{k+1}^2-(x_k+1)x_{k+1}-1=0\implies(x_k+1)=\frac{x_{k+1}^2-1}{x_{k+1}}$$

\begin{align}y_n&=\frac{1}{2}\sum_{k=1}^n\left(\frac{1}{x_k-1}-\frac{1}{x_k+1}\right)\\&=\frac{1}{2}\sum_{k=1}^n\left(\frac{1}{x_k-1}-\frac{x_{k+1}}{x_{k+1}^2-1}\right)\\&=\frac{1}{2}\sum_{k=1}^n\left(\frac{1}{x_k-1}-\frac{1}{2}\left(\frac{1}{x_{k+1}-1}+\frac{1}{x_{k+1}+1}\right)\right)\\&=\frac{1}{4}+\frac{1}{12}-\frac{1}{4}\frac{1}{x_{n+1}-1}-\frac{1}{4}\frac{1}{x_{n+1}+1}+\frac{1}{4}\sum_{k=1}^n\left(\frac{1}{x_k-1}-\frac{1}{x_k+1}\right)\\&=\frac{1}{3}-\frac{1}{4}\frac{1}{x_{n+1}-1}-\frac{1}{4}\frac{1}{x_{n+1}+1}+\frac{1}{2}y_n\\\implies y_n&=\frac{2}{3}-\frac{1}{2}\frac{1}{x_{n+1}-1}-\frac{1}{2}\frac{1}{x_{n+1}-1}\\\implies y_n&\to\frac{2}{3},\quad n\to\infty\end{align}

Convergence is slow, but my Python script indeed evaluates this as $$0.66665666$$ at $$n=100,000$$.