The sum of an infinite geometric series is a remarkable thing... it can be used to prove that 0.99999... (9s forever) is exactly equal to 1.

Given 0.99999... = 0.9 + .09 + .009 + ...

The sum of an infinite geometric&series is defined as sum = t/(1-r), where t is the first term and r is the ratio between terms. Therefore:

1 = 0.9/(1-0.1)

If you don't believe it, ask your math prof.

Series can sometimes be confusing (or at least useful when proving that the sum of an infinite series of positive numbers is negative. or that 0 = 1), so the definition of the value of a series might be helpful. (To be pedantic, the expression "sum of a series" is redundant since a series is a sum).

A series T1 + T2 + T3 + ... = sumi=1 to infinity Ti is said to converge to a value S if the limit S = limn->infinity sumi=1 to n Ti exists. In that case the series is said to be convergent. If no limit exists, it is divergent. Note that the sum up to n can be bounded, even for a divergent series: e.g. 1-1+1-1+1...

Historically, this definition is "backwards", because series were defined first. The modern concept of limits is a later generalisation.

``` inf
---         1
\         ----
/___      ln(n)
n=2
```
Diverges, even though the limn->inf1/ln(n) = 0...

nth-Term Test for Divergence

If a sequence {an} does not converge to 0, then the series Sigma(an) diverges. This does not imply that if the sequence {an} converges to 0 that the series Sigma(an) converges; the test is merely inconclusive. In other words...

``` inf
---         n
\         ----
/___        2
n=1
```
Diverges because limn->inf n/2 = infinity, which does not exist.

Geometric Series

Geometric series are rather simple when it comes to determining their convergence or divergence. First, it would be best to define geometric series.

``` inf
---
\            ar^n = a + ar + ar^2 + ar^3 +... ar^n +..., r != 0
/___
n=0
```
The r term is referred to as the ratio. If 0 < |r| < 1, then the geometric series converges. To determine the series, if it converges, simply follow the formula S = a/(1-r).

Integral Test

This is another rather simple test that works wonderfully for all easily integrable problems. If f is a positive, continuous, decreasing function for n >= 1, and if f(n) = an, then both

``` |\ infinity
|   f(x) dx    and
\| 1

inf
---
\      an
/___
n=0
```
either converge or diverge.

P-Series

P-series are are special series defined as follows:

``` inf
---
\         1/n^p = 1/1^p + 1/2^p +... 1/n^p
/___
n=1
```
To figure out the divergence or convergence of these series, just look at the p. If 0 < p <= 1, then the series diverges. If p > 1, the series converges.

Comparison Tests

If an <= cn for all n, and

``` inf
---
\          cn converges, then
/___
n=1

inf
---
\          an converges
/___
n=1
```
And if cn <= an,
``` inf
---
\            cn diverges, then
/___
n=1

inf
---
\          an diverges
/___
n=1
```
In a similar light, suppose an > 0 and dn > 0. limn-> dn/an = L, where L is finite and positive, then the series
```inf
---
\            dn and
/___
n=1

inf
---
\            an
/___
n=1
```
both either converge or diverge.

There are several other tests for divergence and convergence, but the preceeding tests are a solid foundation for dealing with infinite series.

A geometric series is the sum of terms of the form a*(r^k), where a and r are constants and the sum is over k. The nth partial sum of the geometric series is the sum of a*r^k from k=0 to k=n. We'll call the nth partial sum S(n). So,

S(n)=a+a*r+a*r^2+…+a*r^(n-1)+a*r^n

These sorts of series are useful for analyzing many different, interesting situations. Next, I'll show how to find the value of the nth partial sum and the infinite series. I like this because it's an easy proof, but it shows a really interesting result. One may notice two different ways to get the (n+1)th partial sum (meaning the sum of the geometric series from 0 to n+1).

S(n+1)=a+a*r+…+a*r^n+a*r^(n+1)= S(n)+a*r^(n+1)

You can also get it in a less obvious way, because

S(n+1)=a+(a+a*r+…+a*r^n)*r=a+S(n)*r

Now, you can stick these two equations together by taking one and using it to substitute in for S(n+1) in the other:

S(n)+a*r^(n+1)=a+S(n)*r

Solving for S(n) we get S(n)=a*(1-r^(n+1))/(1-r) to be the value of the nth partial sum, the value of the geometric series from k=0 to k=n. The best thing about this is that to get the value of the infinite series, we just have to take the limit of S(n) as n goes to infinity, provided it exists. The limit exists when |r| < 1. In that case, as n goes to infinity, r^(n+1) goes to zero (because every time you multiply two numbers with |r| < 1 together you get an even smaller one). Thus, for n approaching infinity,

S=a(1-0)/(1-r)=a/(1-r)

Determining the convergence of a series

Because there are only a (relatively) small number of theorems applicable toward determining the convergence of infinite series, it is quite possible to define an "algorithmic" solving process to the question: Does Σun converge/diverge?

Note the use of quotes around the word algorithmic above: it is there to stress the fact I won't be using any formal meta-notation and that I am not putting any claim on fulfilling actual algorithm requirements in this method. It is merely an attempt at organizing all the common tests and tools available to deal with series into a coherent and easy to use list of "if, then" statements.

Anyway, let's say you got an infinite series Σun and want/need/hope to determine whether it is convergent or divergent. Here is probably how you should go at it:

1. Check for Essential Condition on (un)

• if the sequence (un) does not converge toward 0 (no limit or limit different from 0) then Σun diverges

2. Possibly identify Σun as a Special Series

1. Geometric Series: Σarn
2. p-series: Σ1/np (apply p-series theorem)
• if p > 1 then Σ1/np converges
• if 0 p ≤ 1 then Σ1/np diverges (note: includes harmonic series with p = 1)

3. Alternate Series: Σ(-1)nun (where (un) is positive, decreasing and converges to 0) converges

4. Euler Series: Σ1 / n! converges to e (convergence can be proven easily with d'Alembert's ratio test, finding the limit is a whole other problem)

5. Power Series: Σanxn

3. Apply Convergence Tests

1. Absolute Convergence: if Σ | un | converges, then Σun converges (converse is not true).

2. Comparison Test: using a sequence (vn) such as either (un vn) or (un vn) for all n > N (see details in write-up above).

3. Limit Comparison Test (same principle as the Comparison Test, but easier to apply):
• judiciously define a series Σvn such as ( | un / vn | ) converges toward a strictly positive value (0 < lim ( | un / vn | ) < +∞).
• Σun is convergent if and only if Σvn is convergent (see details and example in write-up above).

4. Cauchy Condensation Test:
• if (un) is a decreasing sequence of positive terms.
• then Σun is convergent if and only if Σ2ku2k is convergent.

5. Cauchy Root Test:
• (un) is a sequence of positive terms and l = lim ( (uk)1/k ) (l = ∞, if ( (uk)1/k ) diverges)
• if l < 1 then Σun converges
• if l > 1 (including) then Σun diverges
• if l = 1 then Brian only knows (the test is inconclusive, keep looking)

6. D'Alembert's Ratio Test ("weaker" than Cauchy Root Test but easier to apply):
• (un) is a sequence of positive terms and l = lim (uk+1/uk)
• if l < 1 then Σun converges
• if l > 1 (including) then Σun diverges
• if l = 1 then the test is inconclusive

7. Integral Test: if you can find a positive, continuous, decreasing function f such as f(n) = un for all n ≥ 1 then Σun is convergent if and only if 1+∞f(x)dx is convergent (see details in write-up above)

8. Abel's Test is a sophisticated test mostly used to prove Alternate Series convergence (unlikely to be useful in most cases)

4. If the tests provided in step 3. do not yield any direct results, try splitting the series in a linear combination of series: Σun = λΣan + γΣbn and try again on each of the series

5. If you still cannot conclude, your best chances stand with using the Comparison test and move the problem into solving the convergence of a series Σvn that will be such as either:

• Σvn convergent and un < vn for all n > N
• Σvn divergent and un > vn for all n > N

6. If you still cannot conclude, try banging your head repeatedly against the nearest wall and start again from step 1. (it might work sometimes, provided you do not go too hard on the banging)

7. If you still cannot conclude, consider (in no particular order of preference and depending on possibilities):

1. suicide
2. murder
3. career change
4. turning the page of your textbook

Although I would love to believe that the above list is somewhat close to exhaustive, it is most likely very not so. Out of ignorance, stupidity, plain laziness or a combination of the three, I might have left out useful tests or ideas. If you know of any beneficial addition to this methodology please let me know (try remaining in the strict field of infinite series, though, as I'm not trying to rewrite Principia Mathematica here).

The proof of the formula for the sum of infinite terms of a geometric series, S=a/(1-r), is quite simple.

First, recall the formula for the sum of finite terms of a geometric series, Sn=(a(1-rn))/(1-r)

Replace n with infinity.

Again, recall that r's value can be defined as -1>r>1, and that as fractions are multiplied, their product gets closer to 0. SO, rinfinity is essentially equal to 0, and 1-rinfinity is essentially equal to 1, making a(1-rinfinity) equal to a

Thus, as the term n is no longer needed in the formula (as the quanitiy (1-rinfinity) is equal to 1), niether is n needed in the sum, yielding the formula S=a/(1-r).

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