For any positive, real $k$ there is a $c$ such that $kx + c \gt \ln(x)$ (a proof of this is below using derivatives). If you divide each side by $x$, you can conclude that your function is less than $k + c/x$.
So we have
$$
k + \frac{c}{x} \gt\frac{ \ln(x)}{x} \gt 0
$$
Remember that this is really a load of squeezing inequalities, one for each choice of $k$, with a fitting $c$ to go with it.
In the limit as $x \to \infty$, this becomes
$$
k \geq \lim_{x \to \infty}\frac{\ln(x)}{x} \geq 0
$$
Since $k$ could be any positive, real number, we have the result we want.
Proof of the existence of $c$:
Let's set $x_0 = \frac{1}{k}$. I claim that any $c$ greater than
$$
c' = \ln(x_0) - kx_0
$$
works. We see that $kx + c'$ is tangent to $\ln(x)$ at $x = x_0$, since they have the same functional value and the same derivative. They also do not intersect at any other point since at any point before $x_0$, $\ln(x)$ has greater derivative, and at any point after $kx + c'$ has the greater derivative.
Therefore, any $c\gt c'$ will result in a line $kx + c$ which is strictly greater than $\ln(x)$ for all positive $x$.
Let $\varepsilon > 0$ be given. By the convergence of the series $\sum a_n$ resp. $\sum c_n$, there is an $N(\varepsilon)$ such that for all $N(\varepsilon) \leqslant k < m$ we have
$$\left\lvert\sum_{n=k+1}^m a_n \right\rvert < \varepsilon\land \left\lvert\sum_{n=k+1}^m c_n \right\rvert < \varepsilon.\tag{1}$$
By the inequalities $a_n \leqslant b_n \leqslant c_n$, we have
$$\sum_{n=k+1}^m a_n \leqslant \sum_{n=k+1}^m b_n \leqslant \sum_{n=k+1}^m c_n.$$
Now $(1)$ implies
$$-\varepsilon < \sum_{n=k+1}^m b_n < \varepsilon$$
for $N(\varepsilon) \leqslant k < m$.
Since $\varepsilon$ was arbitrary, the sequence of partial sums is a Cauchy sequence, and $\sum b_n$ converges.
Best Answer
As you pointed out in the question, $(3^n+1)^{\frac{1}{n}}\geq 3$, and on the other hand $$ (3^n+1)^{\frac{1}{n}}\leq (3^n+3^n)^{\frac{1}{n}}=3\cdot 2^{\frac{1}{n}}$$ Since $\lim_{n\to\infty}2^{\frac{1}{n}}=1$, it follows that $\lim_{n\to\infty}(3^n+1)^{\frac{1}{n}}=3$ by the squeeze theorem.