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Chapter 8: Problem 56

Use the test of your choice to determine whether the following series converge. $$\sum_{k=1}^{\infty}\left(1-\frac{1}{k}\right)^{k^{2}}$$

Short Answer

Expert verified
Question: Determine whether the series $$\sum_{k=1}^{\infty}\left(1-\frac{1}{k}\right)^{k^{2}}$$ converges or diverges. Answer: The series converges.

Step by step solution

01

Recall the Root Test

The Root Test can be used to determine the convergence of a series. We apply it by taking the limit of the nth root of the absolute value of the nth term of the sequence. If the limit is less than 1, the series converges. If the limit is greater than 1, the series diverges. If the limit equals 1, the test is inconclusive. Mathematically, it looks like this: Let $$a_n = \left(1-\frac{1}{k}\right)^{k^{2}}$$ be the nth term of the series. Then, we will calculate the limit: $$L = \lim_{n\to\infty}\sqrt[n]{|a_n|}$$
02

Calculate the limit for our given series

We need to calculate the limit: $$L = \lim_{n\to\infty}\sqrt[n]{\left| \left(1-\frac{1}{n}\right)^{n^{2}}\right|}$$ Since $$\left| \left(1-\frac{1}{n}\right)^{n^{2}}\right|$$ is always positive, we can ignore the absolute value: $$L = \lim_{n\to\infty}\sqrt[n]{\left(1-\frac{1}{n}\right)^{n^{2}}}$$ Now, we can rewrite the limit as: $$L = \lim_{n\to\infty}\left(\left(1-\frac{1}{n}\right)^{n^{2}}\right)^{\frac{1}{n}}$$
03

Simplify the exponent

By multiplying the two exponents in the limit, we get: $$L = \lim_{n\to\infty}\left(1-\frac{1}{n}\right)^{n}$$
04

Evaluate the limit

Now, we have a limit that looks like one of the basic limits in calculus: $$L = \lim_{n\to\infty}\left(1-\frac{1}{n}\right)^{n} = \frac{1}{e}$$ Since $$0 < \frac{1}{e} < 1$$, by the Root Test, our given series: $$\sum_{k=1}^{\infty}\left(1-\frac{1}{k}\right)^{k^{2}}$$ converges.

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Most popular questions from this chapter

Consider the geometric series \(S=\sum_{k=0}^{\infty} r^{k}\) which has the value \(1 /(1-r)\) provided \(|r|<1\). Let \(S_{n}=\sum_{k=0}^{n-1} r^{k}=\frac{1-r^{n}}{1-r}\) be the sum of the first \(n\) terms. The magnitude of the remainder \(R_{n}\) is the error in approximating \(S\) by \(S_{n} .\) Show that $$ R_{n}=S-S_{n}=\frac{r^{n}}{1-r} $$

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. $$a_{n}=\sqrt{n} \text { and } b_{n}=2 \ln n, n \geq 3$$

Evaluate the limit of the following sequences or state that the limit does not exist. $$a_{n}=\tan ^{-1}\left(\frac{10 n}{10 n+4}\right)$$

Use the formal definition of the limit of a sequence to prove the following limits. $$\lim _{n \rightarrow \infty} \frac{c n}{b n+1}=\frac{c}{b}, \text { for real numbers } b > 0 \text { and } c > 0$$

An early limit Working in the early 1600 s, the mathematicians Wallis, Pascal, and Fermat were calculating the area of the region under the curve \(y=x^{p}\) between \(x=0\) and \(x=1,\) where \(p\) is a positive integer. Using arguments that predated the Fundamental Theorem of Calculus, they were able to prove that $$\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{k=0}^{n-1}\left(\frac{k}{n}\right)^{p}=\frac{1}{p+1}$$ Use what you know about Riemann sums and integrals to verify this limit.
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