Solutions $a_n$ for $\tan x = x$

Let $\{a_n\}$ be the sequence of consecutive positive solutions to the equation $\tan x = x$. The solutions lie in the intervals $(n\pi, n\pi + \frac{\pi}{2})$ for $n=1, 2, 3, \dots$.

As $n$ becomes large, the solutions $a_n$ approach $n\pi + \frac{\pi}{2}$. We can write $a_n \approx n\pi + \frac{\pi}{2} = (n + \frac{1}{2})\pi$.

We need to determine the convergence of the series $\sum_{n=1}^{\infty} \frac{1}{a_n}$. Using the approximation:

$ \sum_{n=1}^{\infty} \frac{1}{a_n} \approx \sum_{n=1}^{\infty} \frac{1}{(n + \frac{1}{2})\pi} $

This series is comparable to the harmonic series $\sum_{n=1}^{\infty} \frac{1}{n}$. Applying the Limit Comparison Test:

$ \lim_{n\to\infty} \frac{1/(n + \frac{1}{2})\pi}{1/n} = \lim_{n\to\infty} \frac{n}{\pi(n + \frac{1}{2})} = \frac{1}{\pi} \lim_{n\to\infty} \frac{n}{n + \frac{1}{2}} = \frac{1}{\pi} \times 1 = \frac{1}{\pi} $

Since the limit is a positive finite value ($1/\pi$) and the harmonic series $\sum \frac{1}{n}$ diverges, the series $\sum_{n=1}^{\infty} \frac{1}{a_n}$ diverges.

Solutions $b_n$ for $\tan \sqrt{x} = x$

Let $\{b_n\}$ be the sequence of consecutive positive solutions to $\tan \sqrt{x} = x$. Let $y = \sqrt{x}$, so $x=y^2$. The equation transforms to $\tan y = y^2$. Let $\{y_n\}$ be the sequence of consecutive positive solutions for $y$. Then $b_n = y_n^2$.

The solutions $y_n$ are distributed approximately as follows:

• $y_1 \in (0, \frac{\pi}{2})$
• For $n \ge 2$, $y_n \in ((n-1)\pi, (n-1)\pi + \frac{\pi}{2})$

For large $n$, $y_n$ is close to $(n-1)\pi + \frac{\pi}{2} = (n - \frac{1}{2})\pi$. Thus, $y_n \approx (n - \frac{1}{2})\pi$ for $n \ge 2$.

Consequently, $b_n = y_n^2 \approx ((n - \frac{1}{2})\pi)^2$ for $n \ge 2$.

We examine the convergence of the series $\sum_{n=1}^{\infty} \frac{1}{b_n} = \sum_{n=1}^{\infty} \frac{1}{y_n^2}$. We focus on the tail of the series, starting from $n=2$.

$ \sum_{n=2}^{\infty} \frac{1}{y_n^2} \approx \sum_{n=2}^{\infty} \frac{1}{((n - \frac{1}{2})\pi)^2} $

Compare this series to $\sum_{n=2}^{\infty} \frac{1}{n^2}$, which is a convergent p-series ($p=2 > 1$). Using the Limit Comparison Test:

$ \lim_{n\to\infty} \frac{1/((n - \frac{1}{2})\pi)^2}{1/n^2} = \lim_{n\to\infty} \frac{n^2}{\pi^2 (n - \frac{1}{2})^2} = \frac{1}{\pi^2} \lim_{n\to\infty} \left(\frac{n}{n - \frac{1}{2}}\right)^2 = \frac{1}{\pi^2} (1)^2 = \frac{1}{\pi^2} $

Since the limit is a positive finite value ($1/\pi^2$) and $\sum \frac{1}{n^2}$ converges, the series $\sum_{n=2}^{\infty} \frac{1}{y_n^2}$ converges.

As the tail of the series $\sum_{n=1}^{\infty} \frac{1}{b_n}$ converges and the first term $1/b_1$ is finite, the entire series $\sum_{n=1}^{\infty} \frac{1}{b_n}$ converges.

Series Convergence Summary

The analysis shows:

• $\sum_{n=1}^{\infty} \frac{1}{a_n}$ diverges.
• $\sum_{n=1}^{\infty} \frac{1}{b_n}$ converges.

Therefore, the correct option states that $\sum_{n=1}^{\infty} \frac{1}{a_n}$ diverges and $\sum_{n=1}^{\infty} \frac{1}{b_n}$ converges.