Determine whether the series is convergent or divergent.
∞Σ n = 1, 1/(3+e^-n)
To determine the convergence of the series, we can use the limit comparison test with the series 1/n, which is a known divergent series:
lim n→∞ (1/n)/[1/(3+e^-n)]
= lim n→∞ (3+e^-n)/n
= lim n→∞ (3/n + e^-n/n)
= 0 + 0
= 0
Since the limit is finite and positive (0), we conclude that the series ∑(1/(3+e^-n)) converges if and only if the series 1/n converges. Since 1/n diverges, we conclude that the series ∑(1/(3+e^-n)) also diverges.
lim(n→∞) 3+e^-n = 3+0 = 3
so, since a_n→ 1/3, ∑a_n diverges
You are correct, thank you for pointing that out! The limit of 3+e^-n as n approaches infinity does indeed equal 3, and since the terms of the series approach 1/3 as n approaches infinity, the series diverges by the limit comparison test as 1/n also diverges.
To determine whether the series ∑ 1/(3+e^-n) is convergent or divergent, we can use a comparison test.
Let's compare the given series with a known convergent series.
Consider the series ∑ 1/(n^2), where n starts from 1.
Since 1/(3+e^-n) ≤ 1/n^2 for all n ≥ 1, and the series ∑ 1/n^2 is known to be convergent (it is a p-series with p = 2 > 1), we can conclude that the given series is also convergent by comparison.
Therefore, the series ∑ 1/(3+e^-n) is convergent.