27.1018494699 round to nearest hundred-thousandth

The following series are geometric series or a sum of two geometric series. Determine whether each series converges or not. For

Sergeeva-Olga [200]

Answer:

Required solution gives series (a) divergent, (b) convergent, (c) divergent.

Step-by-step explanation:

(a) Given,

$\sum_{n\to 0}^{\infty}\frac{2^n}{9^{2n}+1}$

To applying limit comparison test, let $a_n=\frac{2^n}{9^{2n}+1}$ and $b_n=\frac{9^{2n}}{2^n}$ . Then,

$\lim_{n\to\infty} \frac{a_n}{b_n}=\lim_{n\to\infty}(1+\frac{1}{9^{2n}})=1>0$

Because of the existance of limit and the series $\frac{9^{2n}}{2^n}$ is divergent since $\frac{9^{2n}}{2^n}=(\frac{9^2}{2})^n$ where $\frac{81}{2}>1$ , given series is divergent.

(b) Given,

$\sum_{n\to 1}^{\infty}(\frac{7^n}{7^n+4})$

Again to apply limit comparison test let $a_n=\frac{7^n}{7^n+4}$ and $b_n=\frac{1}{7^n}$ we get,

$\lim_{n\to \infty}\frac{a_n}{b_n}=\frac{1}{7^n+4}=0$

Since $\lim_{n\to \infty} \frac{1}{7^n}=0$ is convergent, by comparison test, given series is convergent.

(c) Given,

$\sum_{n\to 1}^{\infty}\frac{5^n+2^n}{6^n}= \sum_{n\to 1}^{\infty}(\frac{5}{6})^n+\sum_{n\to 1}^{\infty}(\frac{1}{3})^n$ . Now applying Cauchy Root test on last two series, we will get,

\lim_{n\to \infty}|(\frac{5}{6})^n|^{\frac{1}{n}}=\frac{5}{6}=L_1
\lim_{n\to \infty}|(\frac{1}{3})^n|^{\frac{1}{n}}=\frac{1}{3}=L_2

Therefore,

$\lim_{n\to \infty}\frac{5^n+2^n}{6^n}=L_1+L_2=1.16>1$

Hence by Cauchy root test given series is divergent.

5 0

4 years ago

Given the following absolute value function find the range.

arlik [135]

Answer:

Range is (-8,00)

Step-by-step explanation:

4 0

3 years ago

Convert 6.1212 into a rarional number

Ahat [919]

in short, we will start off by making the left-side of the dot and the recurring numbers a variable, say "x", then multiplying it by some power of 10 that moves the recurring numbers over to the left, let's do so

$\bf x = 6.\overline{12}~\hspace{10em} \begin{array}{llll} 100\cdot x&=&612.\overline{12}\\\\ &&606+6.\overline{12}\\\\ &&606+x \end{array} \\\\\\ 100x=606+x\implies 99x=606\implies x = \cfrac{606}{99}\implies x = \cfrac{202}{33}\implies x = 6\frac{4}{33}$