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What is the condition for convergence of the geometric series \(\sum_{k=0}^{\infty} a r^{k} ?\)

Short Answer

Expert verified
Answer: The condition for the convergence of a geometric series is -1 < r < 1, where 'r' is the common ratio between the terms in the series.

Step by step solution

01

Define the geometric series

The given series is a geometric series of the form: \(\sum_{k=0}^{\infty} a r^{k}\), where 'a' is the first term, 'r' is the common ratio between terms (i.e., the multiplier), and 'k' is the index of summation.
02

Recall the formula for the sum of an infinite geometric series

The sum of an infinite geometric series can be given by the formula: \(S = \frac{a}{1-r}\), however this formula is only valid if the series converges.
03

Determine the criterion for convergence

For a geometric series to converge, we must have \(|r| < 1\). This means the common ratio 'r' should be in the open interval of (-1, 1). Based on the information above, the condition for the convergence of the geometric series \(\sum_{k=0}^{\infty} a r^{k}\) is given by: \(-1 < r < 1\).

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Most popular questions from this chapter

Suppose an alternating series \(\sum(-1)^{k} a_{k}\) converges to \(S\) and the sum of the first \(n\) terms of the series is \(S_{n}\) Suppose also that the difference between the magnitudes of consecutive terms decreases with \(k\). It can be shown that for \(n \geq 1,\) $$\left|S-\left[S_{n}+\frac{(-1)^{n+1} a_{n+1}}{2}\right]\right| \leq \frac{1}{2}\left|a_{n+1}-a_{n+2}\right|$$ a. Interpret this inequality and explain why it gives a better approximation to \(S\) than simply using \(S_{n}\) to approximate \(S\). b. For the following series, determine how many terms of the series are needed to approximate its exact value with an error less than \(10^{-6}\) using both \(S_{n}\) and the method explained in part (a). (i) \(\sum_{k=1}^{\infty} \frac{(-1)^{k}}{k}\) (ii) \(\sum_{k=2}^{\infty} \frac{(-1)^{k}}{k \ln k}\) (iii) \(\sum_{k=2}^{\infty} \frac{(-1)^{k}}{\sqrt{k}}\)

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