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Chapter 3: Problem 13

Let \(b \in \mathbb{R}\) satisfy \(0

Short Answer

Expert verified
\(\lim_{n\to\infty} n*b^n = 0\)

Step by step solution

01

Recall the Binomial Theorem

The binomial theorem states that for any real numbers a and b, and a nonnegative integer n, we have \((a+b)^n = a^n + n*a^{(n-1)}*b + ... + a*b^{(n-1)} + b^n\). In this exercise, we're interested when \(a = 1\) and \(b < 1\). Thus the statement becomes \((1+b)^n = 1 + n*b^{(n-1)} + ... + b^n\).
02

Apply the Binomial Theorem to the Exercise

Since \(0 < b < 1\), then for any positive integer n, \(1 < (1 + b)^n\). Using the binomial theorem from Step 1 we get, \(1 < 1 + nb^{n-1} + ... + b^n\). This implies that \(nb^{n-1} < (1 + b)^n - 1\). We also know that \(nb^{n-1} = n*b*b^{(n-1)} = n*b^n\). Therefore, \(n*b^n < (1 + b)^n - 1\). This inequality will be useful in the next step.
03

Show that Limit is Zero

Given \(n*b^n < (1 + b)^n - 1\), as n tends to infinity the expression \((1+b)^n\) will certainly be greater than 1 (on the right side of inequality) whatever the value of \(b\) is, as \(0 < b < 1\). But on the left hand side, \(n*b^n\) will trend towards 0 because \(b^n\) will become very small as \(n\) tends to infinity since \(0 < b < 1\). Thus based on the above inequality, when \(n\) tends to infinity, \(n*b^n ->\) 0. Therefore, we can conclude that \(\lim_{n\to\infty} n*b^n = 0\).
04

Conclusion

We've proven that for any real constant \(b\) where \(0 < b < 1\) and for a sequence in the form \(nb^n\), as \(n\) tends to infinity, the limit of this sequence is 0.

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Most popular questions from this chapter

Give an example of two divergent sequences \(X\) and \(Y\) such that: (a) their sum \(X+Y\) converges, (b) their product \(X Y\) converges.

Let \(\left(x_{n}\right)\) be properly divergent and let \(\left(y_{n}\right)\) be such that \(\lim \left(x_{n} y_{n}\right)\) belongs to \(\mathbb{R}\). Show that \(\left(y_{n}\right)\) converges to 0 .

Let \(x_{1} \geq 2\) and \(x_{n+1}:=1+\sqrt{x_{n}-1}\) for \(n \in \mathbb{N}\). Show that \(\left(x_{n}\right)\) is decreasing and bounded below by 2 . Find the limit.

Show that (a) \(\lim \left(\frac{1}{\sqrt{n+7}}\right)=0\), (b) \(\lim \left(\frac{2 n}{n+2}\right)=2\), (c) \(\lim \left(\frac{\sqrt{n}}{n+1}\right)=0\), (d) \(\lim \left(\frac{(-1)^{n} n}{n^{2}+1}\right)=0\).

Show that the following sequences are not convergent. (a) \(\left(2^{n}\right)\), (b) \(\left((-1)^{n} n^{2}\right)\).
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