use the comparison test to determine whether the series is convergent or divergent
∑ n=1 1/n^n
since ∑ n=1 1/n^2 converges, and n^n > n^2 for n>2, this series also converges, and quite rapidly
To determine whether the series ∑ n=1 1/n^n is convergent or divergent, we can use the Comparison Test.
The Comparison Test states that if 0 ≤ a_n ≤ b_n for all n and if ∑ b_n is convergent, then ∑ a_n is also convergent. Conversely, if 0 ≤ b_n ≤ a_n for all n and if ∑ b_n is divergent, then ∑ a_n is also divergent.
In this case, we want to compare the series ∑ n=1 1/n^n to a known series whose convergence or divergence is already established.
Let's consider the series ∑ n=1 1/n^2. This is a well-known series called the "p-series" with p = 2. It is known that the p-series ∑ 1/n^p is convergent if p > 1 and divergent if p ≤ 1.
Now, we can establish a comparison between our series ∑ n=1 1/n^n and the p-series ∑ n=1 1/n^2:
0 ≤ 1/n^n ≤ 1/n^2 for all n ≥ 1.
Since ∑ 1/n^2 is convergent (p-series with p = 2 and p > 1), by the Comparison Test, it follows that ∑ 1/n^n is also convergent.
Therefore, the series ∑ n=1 1/n^n is convergent.