test the series for convergence or divergence using the alternating series test
the sum from n=1 to infinity of (-1)^n/(3n+1)
I said it converges, is this true?
To test the convergence or divergence of the given series using the Alternating Series Test, we need to check two conditions:
1. The terms of the series must alternate in sign.
2. The absolute value of each term must decrease as n increases.
Let's examine these conditions for the given series:
The terms in the series are given by (-1)^n / (3n+1). Note that (-1)^n alternates between -1 and 1 as n changes.
Now, let's analyze the second condition. we will consider the absolute values of the terms. We have |(-1)^n / (3n+1)| = 1 / (3n+1).
For condition 2, we need to show that the absolute value of each term decreases as n increases. We can do this by taking the limit as n approaches infinity of |a(n+1)| / |an| and check if it is less than or equal to 1.
Let's compute this limit:
lim(n->∞) [ 1 / ((3(n+1)+1) / (3n+1)) ]
lim(n->∞) [ (3n+1) / (3n+4) ] (by replacing n+1 and expanding)
= 1 (by dividing each term by n and applying limit properties)
Since this limit is equal to 1, this satisfies condition 2 of the Alternating Series Test.
Therefore, the given series, the sum from n=1 to infinity of (-1)^n / (3n+1), converges by the Alternating Series Test.