For what values of x does the series 1 + 2^x + 3^x + 4^x + ... + n^x + ... converge?
A. No values of x
B. x >= -1
C. x > -1
D. x < -1
E. All values of x
I think the answer is C, because n^(-1) = 1/n, which diverges.
but what if x were -10
1 + 1/2^10 + 1/3^10 + 1/4^10 .....
Then it would converge. So the series converges for values greater than -1 that are not zero, but it also converges for all values less than -1.
-10 <-1, to the left on a number line
To determine the values of x for which the given series converges, we can use the concept of the p-series test. The p-series test states that a series of the form ∑(n=1 to ∞) n^(-p) converges if and only if p > 1.
In this case, the given series can be written as ∑(n=1 to ∞) n^x. Using the p-series test, we need to find the values of x for which the exponent, x, is greater than 1.
So, the correct answer is:
B. x >= -1
This is because for any value of x greater than or equal to -1, the series will converge. For x less than -1, the series will diverge.