Use Theorem 9.11 to determine the convergence or divergence of the p-series.
1+ 1/6root2 + 1/6root3 + 1/6root4 +...
p=
so some old retired teacher is supposed to know what your theorem 9.11 is?
p series is sum 1 to oo of 1/n^p
if p >1 , converges
if p </= 1 , diverges
here
(1/6) sum 1 to oo of 1/n^.5
p < 1
diverges
Sorry, the Theorem 9.11 is never shown in the work I was given nor explained.
The given series can be written as:
1 + 1/(6√2) + 1/(6√3) + 1/(6√4) + ...
To determine the convergence or divergence of this series, we can use Theorem 9.11, also known as the p-series test. The p-series test states that a series of the form:
∑ (1/n^p)
converges if p > 1 and diverges if p ≤ 1.
In our case, we have:
∑ (1/(6√n))
To apply Theorem 9.11, we need to express the series in the form ∑ (1/n^p). We can do this by rationalizing the denominators:
∑ (1/(6√n)) = ∑ (1/(6 * n^(1/2)))
Now we can see that p = 1/2 in this series.
Since p = 1/2 is smaller than 1, according to Theorem 9.11, the series diverges.
Therefore, the given series 1 + 1/(6√2) + 1/(6√3) + 1/(6√4) + ... diverges.
To determine the convergence or divergence of the given series using Theorem 9.11, we first need to identify the value of p.
The formula for a p-series is given by ∑(1/n^p), where n starts from 1 and goes to infinity.
In the given series, we have 1/6√n for each term. To determine the value of p, we can rewrite 1/6√n as (1/6)*(1/√n) = (1/6)*(n^-1/2).
Comparing this to the formula for a p-series, we can see that p = -1/2.
Now, Theorem 9.11 states:
- If p > 1, the p-series converges.
- If p ≤ 1, the p-series diverges.
Since p = -1/2, which is less than 1, we can conclude that the given series, 1 + 1/6√2 + 1/6√3 + 1/6√4 + ..., is a diverging p-series.