Discuss the convergence and divergence of the series ∑(n=1, infinity)3+cos 𝑛 / 𝑒^𝑛
To determine the convergence or divergence of the series ∑(n=1, ∞) (3 + cos 𝑛) / 𝑒^𝑛, we can use the Limit Comparison Test.
The Limit Comparison Test states that if two series have the same behavior in terms of convergence or divergence, then their terms should have a similar behavior as well. We can apply this test by comparing our series to a known series with easily identifiable behavior.
First, let's consider the series ∑(n=1, ∞) 1 / 𝑒^𝑛. This series is a geometric series with a common ratio of 𝑒^(-1), and we know that a geometric series converges if the absolute value of the common ratio is less than 1. In this case, since 𝑒^(-1) is less than 1, the series ∑(n=1, ∞) 1 / 𝑒^𝑛 converges.
Now, let's proceed to compare the terms of our given series, (3 + cos 𝑛) / 𝑒^𝑛, with the terms of the series 1 / 𝑒^𝑛.
We can rewrite the given series as ∑(n=1, ∞) (3 / 𝑒^𝑛) + (cos 𝑛 / 𝑒^𝑛) and observe the behavior of each term separately.
1. The term 3 / 𝑒^𝑛 is a constant divided by an exponential function. As 𝑒^𝑛 grows larger, the denominator increases rapidly, causing the term to approach zero. Therefore, the series of 3 / 𝑒^𝑛 converges.
2. The term cos 𝑛 / 𝑒^𝑛 involves the cosine function, which oscillates between -1 and 1 as 𝑛 increases. Divided by an exponential function, this term also approaches zero as 𝑛 goes to infinity. Therefore, the series of cos 𝑛 / 𝑒^𝑛 converges.
Since both terms of the given series converge, we can conclude that the original series ∑(n=1, ∞) (3 + cos 𝑛) / 𝑒^𝑛 also converges based on the Limit Comparison Test.