No, he ratio test does NOT imply the convergence for \(\displaystyle{\sum_{{{n}={1}}}^{\infty}}{a}_{{{n}}}\).

asked 2021-06-16

asked 2021-05-05

asked 2021-09-30

asked 2021-06-23

You were asked about advantages of using box plots and dot plots to describe and compare distributions of scores. Do you think the advantages you found would exist not only for these data, but for numerical data in general? Explain.

asked 2021-05-31

a. Plot the graph of f in the viewing window \([0,15]\times [0,10]\).

b. Prove that f is increasing on the interval [0, 15].

asked 2021-08-14

The following advanced exercise use a generalized ratio test to determine convergence of some series that arise in particular applications, including the ratio and root test, are not powerful enough to determine their convergence.

The test states that if

\(\lim_{n \rightarrow \infty} \frac{a_{2 n}}{a_{n}}<1 / 2\)

then \(\sum a_{n}\) converges,while if

\(\lim_{n \rightarrow \infty} \frac{a_{2 n+1}}{a_{n}}>1 / 2\),

then \(\sum a_{n}\) diverges. Let \(\displaystyle{a}_{{{n}}}={\frac{{{1}}}{{{1}+{x}}}}{\frac{{{2}}}{{{2}+{x}}}}\ldots{\frac{{{n}}}{{{n}+{x}}}}{\frac{{{1}}}{{{n}}}}={\frac{{{\left({n}-{1}\right)}!}}{{{\left({1}+{x}\right)}{\left({2}+{x}\right)}\ldots{\left({n}+{x}\right)}}}}\).

Show that \(a_{2 n}/a_{n} \leq e^{-x/2}/ 2\) .

For which x > 0 does the generalized ratio test imply convergence of \(\displaystyle{\sum_{{{n}={1}}}^{\infty}}{a}_{{{n}}}\)?

asked 2021-08-10

Suppose that you want to perform a hypothesis test based on independent random samples to compare the means of two populations. For each part, decide whether you would use the pooled t-test, the nonpooled t-test, the Mann– Whitney test, or none of these tests if preliminary data analyses of the samples suggest that the two distributions of the variable under consideration are
a. normal but do not have the same shape.
b. not normal but have the same shape.
c. not normal and do not have the same shape. both sample sizes are large.