How do you use the ratio test to test the convergence of the series #∑(4^n) /( 3^n + 1)# from n=1 to infinity?

Question

Emily Ammerman · Accepted Answer

Use the ratio test to find that this series diverges...

Let #a_n = 4^n/(3^n+1)# Then: #a_(n+1)/a_n = (4^(n+1)/(3^(n+1)+1))/(4^n/(3^n + 1))# #=(4^(n+1)(3^n+1))/(4^n(3^(n+1)+1))# #=4/3*(1+3^-n)/(1+3^(-n-1))# So #lim_(n->oo) a_(n+1)/a_n = 4/3# Since this is greater than #1#, the series diverges.

Evelyn Agard · Answer

undefined To use the ratio test for the convergence of the series $\sum \frac{4^n}{3^n + 1}$, we compute the limit:

$$
L = \lim_{{n \to \infty}} \left|\frac{a_{n+1}}{a_n}\right|
$$

where $a_n = \frac{4^n}{3^n + 1}$.

First, find $a_{n+1}$ and $a_n$:

$$
a_{n+1} = \frac{4^{n+1}}{3^{n+1} + 1}, \quad a_n = \frac{4^n}{3^n + 1}
$$

Now, compute the ratio:

$$
\frac{a_{n+1}}{a_n} = \frac{\frac{4^{n+1}}{3^{n+1} + 1}}{\frac{4^n}{3^n + 1}}
$$

Simplify:

$$
\frac{a_{n+1}}{a_n} = \frac{4^{n+1}(3^n + 1)}{4^n(3^{n+1} + 1)} = \frac{4(3^n + 1)}{3(3^{n+1} + 1)}
$$

Simplify further:

$$
\frac{a_{n+1}}{a_n} = \frac{4}{3} \cdot \frac{3^n + 1}{3^{n+1} + 1}
$$

As $n$ approaches infinity, $\frac{3^n + 1}{3^{n+1} + 1}$ approaches $\frac{1}{3}$. Therefore,

$$
L = \frac{4}{3} \cdot \frac{1}{3} = \frac{4}{9}
$$

Since $L < 1$, by the ratio test, the series $\sum \frac{4^n}{3^n + 1}$ converges.