Determine whether the p-series is convergent or divergent.
On top of the summation sign (∑) is infinity. Under the summation sign is n=1, and right next to it (to the right of ∑ sign) is 3/n^e.
I was wondering how would you start the problem from start to finish. Any explanations would be appreciated.
Thank you.
since 1/n^p converges if p > 1,
1/n^e converges
May I ask how do you find p?
huh? They told you that p=e
To determine whether the p-series ∑[n=1 to ∞] 3/n^e is convergent or divergent, we need to consider the value of the exponent e.
First, let's go over the p-series. A p-series is of the form ∑[n=1 to ∞] 1/n^p, where p is a positive constant.
The p-series is convergent if the value of p is greater than 1, and it is divergent if the value of p is less than or equal to 1.
In your case, we have an exponent e in the denominator instead of a specific value. We need to determine the values of e for which the p-series converges or diverges.
To do this, we can consider the limit of the series as n approaches infinity, ∑[n=1 to ∞] 3/n^e:
lim(n→∞) 3/n^e
To simplify this expression, we can apply the limit properties:
lim(n→∞) 3/n^e = 3/(lim(n→∞) n^e)
Now, we need to evaluate the limit of n^e as n approaches infinity. There are three cases to consider:
1. If e > 0, then as n approaches infinity, n^e also approaches infinity. In this case, 3/(lim(n→∞) n^e) = 3/∞ = 0. The series is convergent.
2. If e = 0, then n^0 = 1 for all values of n. In this case, 3/(lim(n→∞) n^e) = 3/1 = 3. The series is divergent.
3. If e < 0, then as n approaches infinity, n^e approaches 0. In this case, 3/(lim(n→∞) n^e) = 3/0, which is undefined. The series is divergent.
Therefore, the p-series ∑[n=1 to ∞] 3/n^e converges if e > 0 and diverges if e ≤ 0.