nth root of x is x^1/n
nth root of (n-1)^n is n-1, is it not?
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..
2 answers
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.
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.
1 answer