How do you use the direct comparison test to determine if #Sigma 1/(3^n+1)# from #[0,oo)# is convergent or divergent?

Question

Cameron Alexander · Accepted Answer

The series is convergent

Let #a_n=1/(3^n+1)# We know that #(3^n+1)>3^n# #1/(3^n+1)<1/3^n# Let #b_n=1/3^n# #sum_(n=1)^oob_n=3/2# Geometric series with the common ratio #r=1/3# As the series #sum_(n=1)^oob_n# converges, we can conclude by the comparison test that the series #sum_(n=1)^ooa_n# converges

Emily Ammerman · Answer

undefined To use the direct comparison test for the series $ \sum \frac{1}{3^n + 1} $, we need to find a series whose terms are greater than or equal to the terms of the given series, and whose convergence or divergence is known.

Since $ \frac{1}{3^n + 1} $ is always positive, we can compare it to $ \frac{1}{3^n} $ which is a geometric series with a common ratio $ r = \frac{1}{3} $.

Now, for $ n \geq 1 $, we have:
$$ \frac{1}{3^n + 1} < \frac{1}{3^n} $$

The series $ \sum \frac{1}{3^n} $ converges by the geometric series test because its common ratio is $ \frac{1}{3} $ which is less than 1.

Therefore, by the direct comparison test, since $ \sum \frac{1}{3^n + 1} $ is less than $ \sum \frac{1}{3^n} $, and $ \sum \frac{1}{3^n} $ converges, $ \sum \frac{1}{3^n + 1} $ also converges.