Determine whether the series is convergent, absolutely convergent, conditionally convergent, or divergent.
On top of the summation sign (∑) is infinity. Under the summation sign is n=2, and right next to it (to the right of ∑ sign) is (n root of (n)-1)^n
(Sorry I don't know how to write the n root..(ex. cube root of etc..))
How would you do this problem?
I think I am stuck on the part with the square root..
nth root of x is x^1/n
nth root of (n-1)^n is n-1, is it not?
I think it is (n^(1/n)-1)^n
Using the ratio test,
T(n+1)/Tn
= ((n+1)^(1/(n+1))-1)^(n+1)/(n^(1/n)-1)^n
If you can show that that ratio is less than 1, you're home free.
To determine whether the given series is convergent, absolutely convergent, conditionally convergent, or divergent, we can start by examining the behavior of the terms of the series.
The series is given as ∑(n=2 to infinity) (n^(1/n)-1)^n
Notice that (n^(1/n) - 1)^n can be rewritten as [(n^(1/n))^(n/(1/n)) - 1)^n] = (n - 1)^n.
Now, let's consider the n-th term (n - 1)^n separately.
As n approaches infinity, the term (n - 1)^n behaves like n^n because the negative one is relatively small compared to n.
We can represent the term as n^n.
To evaluate this term, we can use the limit comparison test. Let's choose the geometric series (1/2)^n, which we know is convergent.
Taking the limit as n approaches infinity of the ratio of the n-th terms, we have:
lim as n approaches infinity of [(n - 1)^n / (1/2)^n]
This simplifies to:
lim as n approaches infinity of [(n - 1)^n * (2/1)^n]
Using the properties of limits, we can rewrite this as:
lim as n approaches infinity of [(n - 1)*(2/1)]^n
Now, let's evaluate this limit. We can rewrite it as:
[(n - 1)*(2/1)]^n = [2(n - 1)]^n
Taking the limit of this expression as n approaches infinity, we have:
lim as n approaches infinity of [2(n - 1)]^n
Since the power n approaches infinity and 2(n - 1) is a constant, we can clearly see that this limit approaches infinity.
Therefore, the term (n - 1)^n grows faster than a convergent geometric series as n approaches infinity. Hence, the given series ∑(n=2 to infinity) (n^(1/n)-1)^n is divergent.
In conclusion, the series is divergent.