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Chapter 9: Problem 15

(Geometric series) Show directly that if \(|x|<1\), then \(1 /(1-x)=\sum_{n=0}^{\infty} x^{n}\).

Short Answer

Expert verified
The sum of an infinite geometric series can be expressed as \(1 /(1-x) = \sum_{n=0}^{\infty} x^{n}\) if and only if \(|x| < 1\), which essentially states that the absolute value of 'x' must be less than 1 for the formula to be valid.

Step by step solution

01

Finite Geometric Series

Recall the formula for the sum of a finite geometric series, which is given by \(S = \frac{1 - x^n}{1 - x}\) for \(|x| < 1\). This expression applies when the series has a finite number of terms 'n'. The denominator '1 - x' appears because it is the common ratio of the series.
02

Infinite Geometric Series

If we have an infinite geometric series, we can imagine what happens as 'n' goes to infinity. If \(|x| < 1\), then \(x^n\) will get smaller and smaller as 'n' increases, effectively becoming zero in the limit. So, in the case of an infinite geometric series, we can write \(S = \frac{1 - 0}{1 - x}\). Then, simplifying the numerator, we get \( S = \frac{1}{1 - x}\).
03

Interpreting the Result

So we have found that the sum of an infinite geometric series with ratio 'x' is given by \(\frac{1}{1 - x}\), which is exactly the initial statement to prove, and is valid provided that \(|x| < 1\). This result is crucial in many areas of mathematics and its validity outside the range \(|x| < 1\) is not guaranteed so we have to keep this condition.

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Most popular questions from this chapter

Let \(a_{n}>0\) and suppose that \(\sum a_{n}\) converges. Construct a convergent series \(\sum b_{n}\) with \(b_{n}>0\) such that \(\lim \left(a_{n} / b_{n}\right)=0 ;\) hence \(\sum b_{n}\) converges less rapidly than \(\sum a_{n} .\left[\right.\) Hint \(:\) Let \(\left(A_{n}\right)\) be the partial sums of \(\sum a_{n}\) and \(A\) its limit. Define \(b_{1}:=\sqrt{A}-\sqrt{A-A_{1}}\) and \(b_{n}:=\sqrt{A-A_{n-1}}-\) \(\sqrt{A-A_{n}}\) for \(\left.n \geq 1 .\right]\)

(a) Does the series \(\sum_{n=1}^{\infty}\left(\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}}\right)\) converge? (b) Does the series \(\sum_{n=1}^{\infty}\left(\frac{\sqrt{n+1}-\sqrt{n}}{n}\right)\) converge?

Let \(\left\\{n_{1}, n_{2}, \cdots\right\\}\) denote the collection of natural numbers that do not use the digit 6 in their decimal expansion. Show that \(\sum 1 / n_{k}\) converges to a number less than 80 . If \(\left\\{m_{1}, m_{2}, \cdots\right\\}\) is the collection of numbers that end in 6 , then \(\sum 1 / m_{k}\) diverges. If \(\left\\{p_{1}, p_{2}, \cdots\right.\), ) is the collection of numbers that do not end in 6 , then \(\sum 1 / p_{k}\) diverges.

If the partial sums of \(\sum a_{n}\) are bounded, show that the series \(\sum_{n=1}^{\infty} a_{n} e^{-n t}\) converges for \(t>0\).

(a) If \(\sum a_{n}\) is absolutely convergent and \(\left(b_{n}\right)\) is a bounded sequence, show that \(\sum a_{n} b_{n}\) is absolutely convergent. (b) Give an example to show that if the convergence of \(\sum a_{n}\) is conditional and \(\left(b_{n}\right)\) is a bounded sequence, then \(\sum a_{n} b_{n}\) may diverge.
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