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Does a geometric sum always have a finite value?

Short Answer

Expert verified
Provide a brief explanation. Answer: A geometric sum does not always have a finite value. It converges to a finite value if the common ratio (r) satisfies the condition | r | < 1, and it diverges if | r | >= 1.

Step by step solution

01

Understanding geometric series

A geometric series can be represented by the formula: S_n = a_1 + a_1*r + a_1*r^2 + ... + a_1*r^(n-1) where S_n is the sum of n terms, a_1 is the first term, r is the common ratio, and n is the number of terms.
02

Formula for finite geometric series

For finite geometric series, we can use the formula: S_n = a_1 * (1 - r^n) / (1 - r) This formula is valid only when the common ratio (r) is not equal to 1.
03

Convergence and divergence of geometric series

When considering an infinite geometric series, we are interested in finding the limit of the series as the number of terms, n, approaches infinity: lim_(n->∞) S_n The series converges to a finite value if the limit exists, and diverges otherwise.
04

Criteria for convergence

For an infinite geometric series, the limit of S_n as n approaches infinity can be found using the finite geometric series formula. The series converges to a finite value if: | r | < 1 The limit of the series as n approaches infinity can be computed as: lim_(n->∞) S_n = a_1 / (1 - r)
05

Criteria for divergence

If the common ratio r does not meet the condition | r | < 1, the geometric series diverges and does not have a finite sum. For example, if r = 2, the series becomes: a_1 + 2*a_1 + 4*a_1 + ... The sum of the terms grows without bound as the number of terms increases.
06

Conclusion

A geometric sum does not always have a finite value. It converges to a finite value if the common ratio r is such that | r | < 1, and it diverges if | r | >= 1.

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Most popular questions from this chapter

a. Sketch the function \(f(x)=1 / x\) on the interval \([1, n+1]\) where \(n\) is a positive integer. Use this graph to verify that $$\ln (n+1)<1+\frac{1}{2}+\frac{1}{3}+\dots+\frac{1}{n}<1+\ln n.$$ b. Let \(S_{n}\) be the sum of the first \(n\) terms of the harmonic series, so part (a) says \(\ln (n+1)0,\) for \(n=1,2,3, \ldots\) c. Using a figure similar to that used in part (a), show that $$\frac{1}{n+1}>\ln (n+2)-\ln (n+1).$$ d. Use parts (a) and (c) to show that \(\left\\{E_{n}\right\\}\) is an increasing sequence \(\left(E_{n+1}>E_{n}\right)\). e. Use part (a) to show that \(\left\\{E_{n}\right\\}\) is bounded above by 1 . f. Conclude from parts (d) and (e) that \(\left\\{E_{n}\right\\}\) has a limit less than or equal to \(1 .\) This limit is known as Euler's constant and is denoted \(\gamma\) (the Greek lowercase gamma). g. By computing terms of \(\left\\{E_{n}\right\\}\), estimate the value of \(\gamma\) and compare it to the value \(\gamma \approx 0.5772 .\) (It has been conjectured that \(\gamma\) is irrational.) h. The preceding arguments show that the sum of the first \(n\) terms of the harmonic series satisfy \(S_{n} \approx 0.5772+\ln (n+1)\) How many terms must be summed for the sum to exceed \(10 ?\)

In the following exercises, two sequences are given, one of which initially has smaller values, but eventually "overtakes" the other sequence. Find the sequence with the larger growth rate and the value of \(n\) at which it overtakes the other sequence. $$a_{n}=n^{10} \text { and } b_{n}=n^{9} \ln ^{3} n, n \geq 7$$

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After many nights of observation, you notice that if you oversleep one night, you tend to undersleep the following night, and vice versa. This pattern of compensation is described by the relationship $$x_{n+1}=\frac{1}{2}\left(x_{n}+x_{n-1}\right), \quad \text { for } n=1,2,3, \ldots.$$ where \(x_{n}\) is the number of hours of sleep you get on the \(n\) th night and \(x_{0}=7\) and \(x_{1}=6\) are the number of hours of sleep on the first two nights, respectively. a. Write out the first six terms of the sequence \(\left\\{x_{n}\right\\}\) and confirm that the terms alternately increase and decrease. b. Show that the explicit formula $$x_{n}=\frac{19}{3}+\frac{2}{3}\left(-\frac{1}{2}\right)^{n}, \text { for } n \geq 0.$$ generates the terms of the sequence in part (a). c. What is the limit of the sequence?

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