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A Taylor series usually gives better and better approximations for values of a function the more terms you use. For some series this is true only for certain values of xx. For instance, the series for the natural logarithm,

lnx=n=1(1)n+11n(x1)n\ln x=\sum_{n=1}^{\infty}(-1)^{n+1} \frac{1}{n}(x-1)^n

converges to lnx\ln x only for 0<x20<x \leq 2. If xx is outside this interval of convergence, the series does not converge to a real number. It diverges and thus cannot represent lnx\ln x. In this problem, you will investigate the ratio of a term in this series to the term before it. You will also try to discover a way to find, from this ratio, whether the series converges.

a. The formula for tnt_n, the nnth term in the series for lnx\ln x, is

tn=(1)n+11n(x1)nt_n=(-1)^{n+1} \frac{1}{n}(x-1)^n

Let rnr_n be the ratio tn+1/tn\left|t_n+1 / t_n\right|. Find a formula for rnr_n in terms of xx and nn.

b. Calculate r10r_{10} for x=1.2,x=1.95x=1.2, x=1.95, and x=3x=3.

c. Let rr be the limit of rnr_n as nn approaches infinity. Find an equation for rr in terms of xx.

d. Evaluate rr for x=0.1,x=0,x=0.9x=-0.1, x=0, x=0.9, x=1.9,x=2x=1.9, x=2, and x=2.1x=2.1.

e. Make a conjecture: "The series converges to lnx\ln x whenever the value of xx makes rr- ?and diverges whenever the value of xx makes rr- ?-."

f. If your conjecture is correct, you can use it to show that the series converges if xx is in the interval 0<x<20<x<2. Check your conjecture by showing that it gives this interval.


Answered 2 years ago
Answered 2 years ago
Step 1
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a. First find an expression for tn+1t_{n+1}

tn+1=(1)n+21n+1(x1)n+1 \begin{equation*} t_{n+1}=(-1)^{n+2}\frac{1}{n+1}(x-1)^{n+1} \end{equation*}

now write the ratio as

tn+1tn=(1)n+21n+1(x1)n+1(1)n+11n(x1)nPerform the division=(1)n+2(n+1)nn+1(x1)n+1n\begin{aligned} \left|\frac{t_{n+1}}{t_n}\right|&=\frac{(-1)^{n+2}\frac{1}{n+1}(x-1)^{n+1}}{(-1)^{n+1}\frac{1}{n}(x-1)^n}\text{Perform the division}\\ &=\left|(-1)^{n+2-(n+1)}\frac{n}{n+1}(x-1)^{n+1-n}\right| \end{aligned}

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