CalcII

A geometric series takes the form of Σar^(n-1), where r is the ratio of the growth of the function.

It converges if r < 1

• if r >= 1, the series diverges because the lim -> infin.
• if r = -1, the series diverges because the partial sums alternate between a and 0.

A series that takes the form of 1/n(n+1).

• To utilize this form, you need to use partial fractional decomposition to obtain a usable formula, in this case it is (1/n - 1/(n+1))
• If you look at it like this, every number, except the first term, 1, and the last term, 1/(n+1), cancel each other out.
• So the sum is 1 as n -> ∞

If ΣAn converges, then an -> 0.

• The nth-Term test -
• Σan diverges if lim an fails to exist, or IS NOT ZERO

A harmonic series is a divergent p-series that takes the form of 1/n^p, where p = 1. The series diverges because although the fractions are getting infinitesimally small, there is no upper bound. Thus is continues on to ∞ and diverges.

• As with the harmonic series, the p series takes the form 1/n^p
• Converges if p < 1
• Diverges if p >= 1

If 0 < A < B, for all n, the lesser function (A) either converges or diverges depending on the behavior of the upper bound function.

The direct comparison much be SMALLER than the KNOWN convergent, or LARGER than the known DIVERGENT for this to work.

For two series, A and B, if A > 0 and B > 0, and lim (A/B) = L, where L is finite and POSITIVE, then either both series converge or diverge.

In laymans, take a benchmark of a term that you know to converge or diverge, and compare it to your function in question. They have to be somehow related though, otherwise that's just silly.

IF THE LIMIT IS 0, you learn NOTHING and must try another test!

• Take the integral of the function in question, an improper integral, and compare it with yourbenchmark.
• If f(x) > 0, continuous, and decreasing for all x >= 1, and if An = f(n) then Σan and the integral of f(x) either both converge or diverge.
• The integral test is pretty easy to use

The ratio test looks at the ratio of a term of a series to the immediately preceding term. if in the limit this ratio is less than 1, the series converges, if it is more than 1, the series diverges, and if it equals 1, the TEST IS INCONCLUSIVE.

• Example:
• 3^n / n!
• look at the limit of the ratio of the (n+1) term to the nth term -

• 3^(n+1) / (n + 1)!
• ______________
• 3^n / n!

• Do the basic algebra to rearrange the terms, and you'll find that the factorials cancel out and all you are left with is 3/n+1.
• lim = 0, and since it is less than 1, the series converges!

• Root tests look at limits too, but instead of looking at a ratio of the n+1, you investigate the nth root of the nth term of the series, the result tells the same thing as the results of the ratio test,
• LIM < 1, CONVERGES
• LIM = 1 INCONCLUSIVE!
• LIM > 1 Diverges!

It's a good technique to try if the series involves nth powers.

Many divergent series of positive terms converge if you change the signs of their terms so they alternate. (Alternating harmonic series)

* An alternating series is said to be CONDITIONALLY convergent iff it's convergent as it is but would be divergent if every term were made positive.

* It is ABSOLUTELY convergent if the series remains convergent even when the signs are changed.

Example:

Σ(-1)ⁿ 1/2ⁿ = 1 - 1/2 + 1/4 - 1/8 ...

• if you make the terms positive, you get 1/2ⁿ, which we know is a convergent geometric series.
• Since the latter is convergent, the former must be too, this is what is known as ABSOLUTE convergence.

• Geometric - if a ratio exists where a is the ratio in Σan
• P-series