Math for Non-Geeks/Geometric series

oben Geometric series are series of the form . They are important within several proofs in real analysis. In particular, they are crucial for proving convergence or divergence of other series. We will derive some criteria using them, e.g. the ratio or the root criterion. Geometric sum formula thumb| A video explaining the geometric series (in German).(YouTube-Video by the channel Quatematik) We now prove the geometric sum formula for partial sums of the geometric series. Geometric sum formula For all real and for all there is: We have Geometric series thumb|250px|A (German) song about the geometric series (Youtube-Video by DorFuchs) thumb|250px|The geometric series converges for , or . We consider two cases: and . Case We consider the geometric series for any , which especially means . The sum formula above applies to the partial sums in that case: So the geometric series converges if and only if the sequence of partial sums converges. This is the case if and only if converges. We know that converges to if and only if and it converges to , if and only if . In this section, we only care about the first case of convergence: If , then the geometric series converges. Now, let us determine its limit: Alternatively, convergence for can be shown directly, using the definition. alternative proof that the geometric series converges Prove that the geometric series with converges to . We need to show that for each there is an , such that for all The geometric sum formula yields Since the geometric sequence with converges to 0, so does . Hence, for every there is an with for all Since is constant, there is an with for all This implies the desired convergence. Let be given. The geometric sequence with converges to 0. Since , for a given there is an with for all With the geometric sum formula, this implies for all that The limit of the series is hence: . Case For , we have for all , that . Therefore, the sequence cannot converge to 0. So the series must diverge (this argument is called term test and will be considered in detail, later) The divergence becomes particularly obvious, if is positive, e.g. for . In this case, for all , we have and may estimate the partial sums: So the sequence of partial sums is bounded from below by the sequence , which in turn diverges to . So the series must diverge, as well. Conclusion We have learned: for , and , the geometric series diverges. These three cases can be concluded into one case . However, if , then the geometric series converges to : geometric series The geometric series converges if and only if . In that case, the limit is , or written in shorthand notation: For , and there is Example problems Problem 1 problems: geometric series Compute the values (=limits) of the following series: thumb| in a picture Careful! The series starts with and not . So we need to shuffle terms around first before we can apply the sum formula with : This series starts with . We perform an index shift to get it starting with : Problem 2 geometric series with special Let with and . Find the limits of the following six geometric series: and and and and and and Problem 3 index shifting Let with . Find the limits of the three geometric series for For and there is Problem 4 Sequences which relate to the geometric series Solve the following three problems: For all real and , prove the limit . For all with , prove the limit . Find the limits of the series and . For all and we have to establish the left-hand side is re-arranged as follows: In the chapter examples for limits , we proved that holds for (reason: grows exponentially and "beats" the linearly growing ). The limit theorems hence yield and . Therefore We re-use the solution to sub-exercise 2 with : The second limit also follows from , using an index shift: Alternative proof to sub-exercise 1 We may also reconstruct the limit similarly to the geometric sum formula: There is a factor of in the denominator, so applying the trick "multiply by and subtract" twice, we should get to the result. We start from And multiply the equation by : Next, both results are subtracted: On the left side, we may factor out a second : Alternative proof to sub-exercise 3 We may also add and subtract a 1: Analogously to sub-exercise 3, one may show for every that: just replace by unten