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 $x$. For instance, the series for the natural logarithm,

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

converges to $\ln x$ only for $0<x \leq 2$. If $x$ is outside this interval of convergence, the series does not converge to a real number. It diverges and thus cannot represent $\ln x$. 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 $t_n$, the $n$th term in the series for $\ln x$, is

$t_n=(-1)^{n+1} \frac{1}{n}(x-1)^n$

Let $r_n$ be the ratio $\left|t_n+1 / t_n\right|$. Find a formula for $r_n$ in terms of $x$ and $n$.

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

c. Let $r$ be the limit of $r_n$ as $n$ approaches infinity. Find an equation for $r$ in terms of $x$.

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

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

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

Solution

Verified**a.**
First find an expression for $t_{n+1}$

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

now write the ratio as

$\begin{aligned} \left|\frac{t_{n+1}}{t_n}\right|&=\frac{(-1)^{n+2}\frac{1}{n+1}(x-1)^{n+1}}{(-1)^{n+1}\frac{1}{n}(x-1)^n}\text{Perform the division}\\ &=\left|(-1)^{n+2-(n+1)}\frac{n}{n+1}(x-1)^{n+1-n}\right| \end{aligned}$

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