Let $\{x_{n}\}$ be a sequence of non-negative real numbers such that $x_{n + 1}^2 =6x_{n} +7$ for all $n ≥ 2$
Which one of the following is true?
$\quad(A).\space$ If $x_{2} > x_{1} > 7$ then $\displaystyle\sum_{n=1}^{\infty} x_{n}$ is convergent
$\quad(B).\space$ If $x_{2} > x_{1} > 7$ then $\displaystyle\sum_{n=1}^{\infty} \dfrac{(-1)^{n}}{x_n} $ converges
$\quad(C).\space$ If $7 > x_{2} > x_{1} > 6$ then $\displaystyle\sum_{n=1}^{\infty} \dfrac{1}{x_n}$ is convergent
$\quad(D).\space$ None of these is convergent
I tried by finding the infinite term which comes to be 7 for all x so according to me any summation will never converge as the infinite term is not approaching 0. but the answer is given as (B)
It follows from the recurrence relation $x_{n + 1}^2 =6x_{n} +7$ that $(x_n)$ does not converge to zero, that excludes (A).
Similarly, $y_n = 1/x_n$ satisfies the recurrence relation $1 + 6 y_n + 7 y_n^2$, so that $(y_n)$ does not converge to zero either. That excludes (B) and (C).
It is also correct that $(x_n)$ converges to $7$, but the exact limit is actually not needed to exclude the convergence of the series in (A), (B), or (C), only that $(x_n)$ resp. $(1/x_n)$ does not converge to zero.