determine if this converges and methods used to find its covergence or divergence.
The series as n = 1 to infinity of [(2n+3)/(3n+2)]^n
To determine the convergence or divergence of the given series, we can use the comparison test. Let's break down the steps:
Step 1: Simplify the given series.
The series can be written as:
[(2n+3)/(3n+2)]^n = [(2 + 3/n)/(3 + 2/n)]^n
Step 2: Choose a comparable series.
We can choose the series (1/n)^n as a comparable series because the powers and the coefficients have similar forms.
Step 3: Apply the comparison test.
Comparing the given series, [(2 + 3/n)/(3 + 2/n)]^n, with the comparable series, (1/n)^n, we can take the limit of their ratio as n approaches infinity.
Let's calculate the limit:
lim(n->∞) [(2 + 3/n)/(3 + 2/n)]^n / (1/n)^n
We can take the reciprocal of the limit:
lim(n->∞) [(3 + 2/n)/(2 + 3/n)]^n * n^(-n)
Now, we can simplify the expression within the limit using algebraic manipulations:
lim(n->∞) [(3 + 2/n)*(n + 3)/(2 + 3/n)]^n * n^(-n)
= lim(n->∞) [(3n + 9)/(2n + 6)]^n * n^(-n)
Step 4: Evaluate the limit.
Taking the limit as n approaches infinity, we consider the highest powers in the numerator and denominator to determine the behavior of the limit without getting a specific value.
lim(n->∞) [(3n + 9)/(2n + 6)]^n * n^(-n)
= [lim(n->∞) (3n + 9)/(2n + 6)]^∞ * [lim(n->∞) n^(-n)]
We can evaluate each limit separately:
[lim(n->∞) (3n + 9)/(2n + 6)]^∞ = [3/2]^∞ (a constant)
lim(n->∞) n^(-n) can be evaluated by taking the natural logarithm:
ln[lim(n->∞) n^(-n)] = lim(n->∞) [-n*ln(n)]
= -∞ (since the logarithm of n goes to negative infinity faster than n itself approaches infinity)
So, the first part becomes [3/2]^∞ and the second part goes to -∞.
Step 5: Determine convergence or divergence.
If the first part, [3/2]^∞, is finite (not zero) and the second part goes to -∞, the entire expression is convergent.
In this case, [3/2]^∞ is a finite positive number since 3/2 is greater than 1. However, the second part, -∞, indicates an infinitely small value.
Therefore, the given series, [(2n+3)/(3n+2)]^n, is convergent.
To summarize, the convergence of the series was determined using the comparison test and evaluating the limits involved. Based on the calculations, the series [(2n+3)/(3n+2)]^n is convergent.