For what value(s) of r is the series the summation from n=1 to infinity of 1/(n^r+3) convergent?
a) r > -2
b) r > -1
c) r ≥ -2
d) All values of r
Please can you show me your work with the answer? Thank you very very much
recall that 1/n^p converges for p>1
PLEASE, I don't get it. Which one is the answer?
huh? really?
For what values of r is r+3 > 1 ??
r>-2. thanks
To determine the convergence of the series, we can use the p-series test, which states that the series of the form ∑(1/n^p) is convergent if p > 1 and divergent if p ≤ 1.
In this case, our series is: ∑(1/(n^r+3))
To apply the p-series test, we need to rewrite the series in the form of ∑(1/n^p). Hence, we need to manipulate the expression in such a way that it matches the p-series.
Let's separate the series into two parts:
∑(1/n^r) + ∑(1/n^3)
The first part, ∑(1/n^r), resembles the p-series form. According to the p-series test, this part will be convergent if r > 1.
The second part, ∑(1/n^3), is a known convergent series. It doesn't depend on the value of r since the power is fixed at 3.
To determine the convergence of the entire series, we need both parts to be convergent.
From the first part, we know that r > 1 for convergence, and the second part is always convergent.
Therefore, the series ∑(1/(n^r+3)) is convergent when r > 1.
Now let's evaluate the answer choices:
a) r > -2: Since -2 < 1, this is not true.
b) r > -1: Similarly, -1 < 1, so this is also not true.
c) r ≥ -2: This choice includes the values of r that are greater than or equal to -2. Since -1 is included, this choice is incorrect as well.
d) All values of r: This is not correct either, as we have shown that r > 1 for convergence.
In conclusion, the correct answer is none of the given choices. The series is only convergent when r > 1.