Calculus 3 Convergence and Divergence

Use the comparison test to determine the convergence or divergence of the series, showing all work.

4 + 1/5 + 0.3 + 1/(3+√2) + 1/(9+√3) + 1/(27+√4) + 1/(81+√5) + …

The comparison test can be used when Σan is a series with no negative terms.

Σan will converge if there is a convergent series Σcn where an ≤ cn

Σan will diverge if there is a divergent series Σdn where an ≥ dn

The series 4 + 1/5 + 0.3 + 1/(3+√2) + 1/(9+√3) + 1/(27+√4) + 1/(81+√5) + … has the nth term 1/(3n + √n-1)

The first three terms will be ignored.

Then the series will be compared to the converging geometric series Σ∞n=1 1/3n

The terms of 1/(3n + √n-1) are smaller than the terms of 1/3n and so this agrees with the concept that Σan will converge if there is a convergent series Σcn where an ≤ cn

With a term by term comparison we can see that an ≤ cn

1/(3 + √2) + 1/(9+√3) + 1/(27+√4) + 1/(81+√5) + … ≤ 1/3 + 1/9 + 1/27 + 1/81 + …

So Σ∞n=1 1/(3n + √n-1) converges by comparison with Σ∞n=1 1/3n

Use the alternating series test to show the convergence or divergence of the series Σ∞i=1(-1)(i+1) (i+3)/(i2 + 10), showing all work.

In the alternating series test the series has values alternating from positive to negative or from negative to positive. The part (-1)(i+1) shows that this is the case for this series.

With the alternating series test the series Σ∞n=1 (-1)n+1 an will converge if,

a. an > 0, in this case an = (i+3)/(i2 + 10)

b. a1 > a2 > a3 >… , this means the terms in the series are decreasing. The terms for the given series in this problem are 4/11 - 5/14 + 6/19 …, this implies that each term in this case is decreasing.

c. limn→∞ an = 0, with our given series limn→∞ (i+3)/(i2 + 10) = ∞/∞ which is an indeterminate form. We can use L’Hôpital’s rule to find the solution which states that the limit of the differential of a function is the limit of the original function. If f(i) = i+3 and g(i) = i2 + 10 then f’(i)/g’(i) = 1/i and limn→∞ 1/i...