Why is the limit of the sum n=1 to infinity of ln(2(n+1))–ln(2n) divergent? I thought it would be -ln2 because all the other terms cancel out cancel out.
L'Hopitals rule:
(1/2n+2)(2n/1)
2n/(2n+2)
second application of the rule:
2/2=1
so at infinity, you are adding 1 + 1+ ...
which means it is divergent.
To determine whether the series ∑(n=1 to ∞) ln(2(n+1)) - ln(2n) is convergent or divergent, we can analyze the behavior of the individual terms and apply a convergence test. In this case, we can simplify the expression before proceeding.
The given expression ∑(n=1 to ∞) ln(2(n+1)) - ln(2n) can be rewritten as a telescoping series:
∑(n=1 to ∞) ln(2(n+1)) - ln(2n)
= ln(3) - ln(2) + ln(4) - ln(3) + ln(5) - ln(4) + ln(6) - ln(5) + ...
When we look for canceling terms, it might seem like everything cancels out, giving us a series of 0's:
(ln(3) - ln(2)) + (ln(4) - ln(3)) + (ln(5) - ln(4)) + (ln(6) - ln(5)) + ...
However, we need to remember that we are considering the sum from n = 1 to infinity. As n approaches infinity, the number of terms in parentheses grows without bound.
Each individual term (ln(n+1) - ln(n)) is a difference between natural logarithms of consecutive terms, resulting in ln((n+1)/n) for each term. However, ln((n+1)/n) is not equal to zero. Instead, it simplifies to ln(1 + (1/n)).
As n approaches infinity, ln(1 + (1/n)) approaches ln(1) = 0. Therefore, the individual terms in the series approach zero as n grows, but they never actually become zero.
Now, let's use the limit comparison test to analyze the convergence of the series. We will compare this series with a known convergent series.
Consider the series ∑(n=1 to ∞) ln(1/n). This is a known divergent series, meaning it goes to negative infinity.
Using the limit comparison test, we can take the limit as n approaches infinity of the ratio of the terms in the original series to the terms in the series ln(1/n):
lim(n->∞) [ln(1 + 1/n)] / [ln(1/n)]
Using L'Hôpital's rule, we can evaluate this limit:
lim(n->∞) [(1/n) / (1 + 1/n)] / (1 / (n * ln(1/n)))
= lim(n->∞) [(1/n) / (1 + 1/n)] / (1 / (n * (ln(n) - ln(1))))
= lim(n->∞) [(1/n) / (1 + 1/n)] / (1 / (n * ln(n)))
Simplifying further:
= lim(n->∞) (1 + 1/n) / ln(n)
As n approaches infinity, the numerator approaches 1, and the denominator approaches infinity. Therefore, the overall limit equals 0.
Since the limit of the ratio of the terms in the original series to the terms in the divergent series is 0, we can conclude that the original series ∑(n=1 to ∞) ln(2(n+1)) - ln(2n) is also divergent.
In summary, the series is divergent because, even though the individual terms approach 0, the number of terms in the series is infinite, and their sum does not converge to a finite value.