Discuss the convergence and divergence of the series ∑n=1 to ∞ 3+cos n/e^n
well, ∑1/e^n converges
and |cos n| ≤ 1
so what does that tell you?
can i get full solution for this question
is it divergence
To determine the convergence or divergence of the series ∑n=1 to ∞ (3 + cos(n))/e^n, we can use various tests for convergence. Let's go through some popular methods:
1. Ratio Test:
The ratio test is commonly used for series involving exponentials. For the given series, let's calculate the ratio of successive terms:
lim(n→∞) ((3 + cos(n))/e^n+1) / ((3 + cos(n))/e^n)
= lim(n→∞) (3 + cos(n))/(3 + cos(n+1)) * e
= e
Since the limit is e (>1), the series diverges by the ratio test.
2. Comparison Test:
We can compare the given series to a known convergent or divergent series. In this case, let's compare it to the geometric series ∑n=1 to ∞ (1/e)^n:
Since (1/e)^n is a geometric series with a common ratio less than 1, it converges. Now, comparing the given series:
0 ≤ (3 + cos(n))/e^n ≤ 1/e^n
Since the geometric series ∑n=1 to ∞ 1/e^n converges, and the terms of the given series are bounded above and below by this convergent series, the given series ∑n=1 to ∞ (3 + cos(n))/e^n also converges.
3. Absolute Convergence:
To determine absolute convergence, we can investigate the convergence of the absolute value of each term in the series. The absolute value of (3 + cos(n))/e^n is:
| (3 + cos(n))/e^n | = (|3 + cos(n)|)/e^n
For |3 + cos(n)|, the maximum value is 4, as the cosine function oscillates between -1 and 1. Hence,
(3 + cos(n))/e^n ≤ 4/e^n
Since ∑n=1 to ∞ 4/e^n is a geometric series that converges, the given series ∑n=1 to ∞ (3 + cos(n))/e^n also converges absolutely.
Therefore, we conclude that the series ∑n=1 to ∞ (3 + cos(n))/e^n converges.