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

Question

Emily Ammerman · Accepted Answer

The series is convergent and:

#sum_(n=1)^(+oo) (3/4)^n = 3# The ratio to test is: #r= a_(n+1)/a_n= frac ((3/4)^(n+1)) ((3/4)^n) =3/4# As #r<1# the series is convergent. We can note that this is a particular case of the geometric series: #sum_(n=0)^(+oo) x^n = 1/(1-x)# for #|x|<1#. So that we can also calculate the sum: #sum_(n=1)^(+oo) (3/4)^n = -1 + sum_(n=0)^(+oo) (3/4)^n = -1+1/(1-3/4) = -1 + 1/(1/4)=-1+4=3 #

Isabella Ackerman · Answer

undefined To use the ratio test to test the convergence of the series $\sum_{n=1}^\infty \left(\frac{3}{4}\right)^n$, you need to compute the limit of the ratio of consecutive terms as $n$ approaches infinity.

The general form of the ratio test states: If $L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|$, then:
- If $L < 1$, the series converges absolutely.
- If $L > 1$, the series diverges.
- If $L = 1$, the test is inconclusive.

For the given series, $a_n = \left(\frac{3}{4}\right)^n$, so $a_{n+1} = \left(\frac{3}{4}\right)^{n+1}$.

Now, compute the limit:
$$L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{\left(\frac{3}{4}\right)^{n+1}}{\left(\frac{3}{4}\right)^n} \right|$$
$$= \lim_{n \to \infty} \left| \frac{\left(\frac{3}{4}\right)^{n}\cdot\frac{3}{4}}{\left(\frac{3}{4}\right)^n} \right|$$
$$= \lim_{n \to \infty} \left| \frac{\frac{3}{4}}{1} \right|$$
$$= \frac{3}{4}$$

Since $L = \frac{3}{4} < 1$, by the ratio test, the series $\sum_{n=1}^\infty \left(\frac{3}{4}\right)^n$ converges absolutely.