Does the series converge or diverge? Use the Root test
((1/n)-(1/n^2))^n
To determine if the series converges or diverges using the Root Test, we need to find the limit of the absolute value of the nth root of the terms of the sequence as n approaches infinity.
First, let's rewrite the series in terms of n:
((1/n)-(1/n^2))^n = (1/n^n - 1/n^2n)^n = (n^n)^(-1/n) * (1 - 1/n^2)^n
Now, let's find the limit of the absolute value of both factors as n approaches infinity.
The limit of (n^n)^(-1/n) as n approaches infinity can be found by taking the logarithm of it:
ln((n^n)^(-1/n)) = (-1/n) * ln(n^n) = (-1/n) * n * ln(n) = -ln(n)
As n approaches infinity, -ln(n) approaches negative infinity.
Now, let's consider the limit of (1 - 1/n^2)^n as n approaches infinity:
ln((1 - 1/n^2)^n) = n * ln(1 - 1/n^2)
Using the fact that ln(1 + x) ≈ x for small x, we can approximate ln(1 - 1/n^2) as -1/n^2:
n * ln(1 - 1/n^2) ≈ n * (-1/n^2) = -1/n
As n approaches infinity, -1/n approaches 0.
Now, let's combine the limits of both factors:
lim[(n^n)^(-1/n) * (1 - 1/n^2)^n] = lim[(n^n)^(-1/n)] * lim[(1 - 1/n^2)^n] = (-∞) * 0 = 0
Since the limit is 0, the Root Test indicates that the series converges.
Therefore, the series ((1/n)-(1/n^2))^n converges.