Convergence of $\{a_n\}$ given that $b_n = a_n + 1/a_n$ converges

Question

I am given a sequence $\{a_n\}$ with $a_n > 0$ and $b_n = a_n + 1/a_n$.

I am first asked to assume that $a_n \ge 1$ and show that the convergence of $\{b_n\}$ implies the convergence of $\{a_n\}$.

My argument is:

Since $\{b_n\}$ converges, $\{b_n\}$ is Cauchy, hence for every $\epsilon > 0$ some $N$ such that for $n_i,n_j > N$, $|b_{n_i} - b_{n_j}| < \epsilon$. Then $|b_{n_i} - b_{n_j}| = |a_{n_i} - a_{n_j} + 1/a_{n_i} - 1/a_{n_j}| < \epsilon$. Suppose that $a_{n_i} = a_{n_j}$. Then clearly $|a_{n_i} - a_{n_j}| < \epsilon$. Alternatively, suppose without loss of generality that $a_{n_i} > a_{n_j}$. Then $1/a_{n_i} - 1/a_{n_j} < 0$ while $a_{n_i} - a_{n_j} > 0$. It is straightforward to observe that $-(1/a_{n_i} - 1/a_{n_j}) \le a_{n_i} - a_{n_j}$ because of the condition $a_n \ge 0$, so it follows that $0 < |a_{n_i} - a_{n_j}| < |a_{n_i} - a_{n_j}|$, so then $|a_{n_i} - a_{n_j}| < |a_{n_i} - a_{n_j} + 1/a_{n_i} - 1/a_{n_j}| < \epsilon$, hence $\{a_n\}$ is Cauchy and therefore converges.

However, I am next asked to assume that $\{b_n\}$ converges but only that $a_n > 0$ and demonstrate, through construction of a counterexample, that it does not necessarily follow that $\{a_n\}$ converges. I am not entirely sure how to do this.

Answer 1

Hint: draw a graph of the function $f(x)= x + \frac{1}{x}, x>0$, what are solutions of $f(x) = b$? with $b = \dfrac{5}{2}$ for example

Answer 2

You are really overthinking the second part.

Take the sequence $a_n = a, 1/a, a, 1/a, a, 1/a, ... $, then $b_n = a + 1/a$ and the sequence $b_n$ is convergent, but $a_n$ is not convergent unless a = 1.

Answer 3

Let $$a_n=2+\sqrt 3(-1)^n$$which is not convergent while $b_n=4$ is convergent to $4$.