Does the series converge or diverge? Use the comparison test. Thank you
((1/n)-(1/n^2))^n
a simple convergent geometric series is r^n where r<1.
(1/n - 1/n^2) = (n-1)/n^2 < 1 for all n>0
In fact,
1/n (1 - 1/n) < 1/n < 1/2 for n>2.
So, An < 1/2^n for n>2
To determine whether the series converges or diverges, we can use the comparison test.
The first step is to find a simpler series whose convergence or divergence is known. In this case, let's compare the given series with the series (1/n)^n.
Applying the limit comparison test, we take the limit as n approaches infinity of the ratio between the two series:
lim (n approaches infinity) [((1/n)-(1/n^2))^n / (1/n)^n]
Simplifying further, we get:
lim (n approaches infinity) [(1 - 1/n)^n * n^n / n^n]
Notice that (1 - 1/n)^n approaches e^(-1) as n approaches infinity, by the definition of the exponential function e^x.
So, the limit becomes:
lim (n approaches infinity) [e^(-1)]
Since e^(-1) is a positive constant, the series (1/n)^n and the given series have the same convergence behavior.
Now, we can determine the convergence or divergence of (1/n)^n. The series (1/n)^n is a p-series with p = 1, which means it is a harmonic series. The harmonic series diverges.
Therefore, using the comparison test, we conclude that the given series ((1/n)-(1/n^2))^n also diverges.