Question

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(1 point) Book Problem 27 Determine whether the following sequences are divergent or convergent. If convergent, evaluate the limit. If divergent to infinity, state your answer as "INF" (without the quotation marks). If divergent to negative infinity, state your answer as "MINF". If divergent without being infinity or negative infinity, state your answer as "DIV". Sequence on = ln(9m² +1) – In(nº +1), lim -

Answer 1

27) Given, $a_n=\ln(9n^2+1)-\ln(n^2+1)$

i.e., $a_n=\ln\left(\frac{9n^2+1}{n^2+1}\right)$

Now, $\lim_{n\to\infty}a_n=\lim_{n\to\infty}\ln\left(\frac{9n^2+1}{n^2+1}\right)$

i.e., $\lim_{n\to\infty}a_n=\lim_{n\to\infty}\ln\left(\frac{(9n^2+1)/n^2}{(n^2+1)/n^2}\right)$

i.e., $\lim_{n\to\infty}a_n=\lim_{n\to\infty}\ln\left(\frac{9+(1/n^2)}{1+(1/n^2)}\right)$

i.e., $\lim_{n\to\infty}a_n=\lim_{n\to\infty}\ln\left(\frac{9+(1/n)^2}{1+(1/n)^2}\right)$

i.e., $\lim_{n\to\infty}a_n=\lim_{k\to0}\ln\left(\frac{9+k^2}{1+k^2}\right)$ [Taking (1/n) = k]

i.e., $\lim_{n\to\infty}a_n=\ln\left(\frac{9+0^2}{1+0^2}\right)$

i.e., $\lim_{n\to\infty}a_n=\ln\left(\frac{9+0}{1+0}\right)$

i.e., $\lim_{n\to\infty}a_n=\ln\left(\frac{9}{1}\right)$

i.e., $\lim_{n\to\infty}a_n=\ln9$

Therefore, the given sequences is convergent.

And, $\mathbf{\lim_{n\to\infty}a_n=\ln9}$ .

29) Given, $a_n=1000(1.03)^n$

(a) Now, $a_1=1000(1.03)^1$

i.e., $a_1=1000\times1.03$

i.e., $\mathbf{a_1=1030}$

and, $a_2=1000(1.03)^2$

i.e., $a_2=1000\times1.0609$

i.e., $\mathbf{a_2=1060.9}$

and, $a_3=1000(1.03)^3$

i.e., $a_3=1000\times1.092727$

i.e., $a_3=1092.727$

i.e., $\mathbf{a_3\approx1092.73}$

and, $a_4=1000(1.03)^4$

i.e., $a_4=1000\times1.12550881$

i.e., $a_4=1125.50881$

i.e., $\mathbf{a_4\approx1125.51}$

(b) Here, $\lim_{n\to\infty}a_n=\lim_{n\to\infty}1000(1.03)^n$

i.e., $\lim_{n\to\infty}a_n=1000\times\lim_{n\to\infty}(1.03)^n$

i.e., $\lim_{n\to\infty}a_n=1000\times(1.03)^{\infty}$

i.e., $\lim_{n\to\infty}a_n=1000\times\infty$

i.e., $\lim_{n\to\infty}a_n=\infty$

This implies that the sequence is divergent.

Therefore, enter DIV.