Determine whether the series is convergent or divergent. ∑from n = 1 to∞. 1/ (8 + e^−n)
well, the terms of the sequence approach 1/8, right?
so what about ∑ 1/8 ?
no, what about the e^-n?
To determine whether the series ∑ (1 / (8 + e^(-n))) is convergent or divergent, we can use the comparison test.
First, we need to find a series whose behavior is known. Let's compare it with another series ∑ (1 / (8 + 1/n)), which is easier to handle since the exponential term is removed.
Now, we can compare the two series by taking the limit of their terms as n approaches infinity:
lim (n→∞) (1 / (8 + e^(-n))) / (1 / (8 + 1/n))
= lim (n→∞) (8 + 1/n) / (8 + e^(-n))
= 8 / 8
= 1
Since the limit of the terms is finite and nonzero (1 in this case), it means that the convergence or divergence of the original series is the same as the series ∑ (1 / (8 + 1/n)).
Now, we can analyze the series ∑ (1 / (8 + 1/n)). This is a sum of terms of the form 1 / (8 + 1/n). As n approaches infinity, the denominator of each term approaches 8, and the value of each term approaches 1/8.
We can say that the series ∑ (1 / (8 + 1/n)) is a harmonic series with a constant term of 1/8. The harmonic series ∑ (1 / n) is known to be divergent, so scaling it by a constant (in this case, 1/8) does not change its divergence. Thus, the series ∑ (1 / (8 + 1/n)) is also divergent.
Therefore, by the comparison test, the original series ∑ (1 / (8 + e^(-n))) is also divergent.