Mathematics, 04.02.2020 16:57 alex34403

Convergence/divergence of the sums of series!

problems attached. walk me through them?

for 8, i started by using the ratio test.

| 2^n+1 / x^n+1 | / | 2^n/ x^n | = 2/x

where do i go from here?

9. can i also use the ratio test here?

10. and 11. no clue.

Convergence/divergence of the sums of series! problems attached. walk me through them?

Answers: 3

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Answer from: janinecastillo01

The ratio test says that the series $\sum a_n$ converges if the absolute value of the ratio between consecutive terms approaches a number $0\le L$ , and diverges otherwise:

$\displaystyle\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|$

You have

$\displaystyle\lim_{n\to\infty}\left|\frac{\frac{2^{n+1}}{x^{n+1}}}{\frac{2^n}{x^n}}\right|=\lim_{n\to\infty}\left|\frac{2^{n+1}}{2^n}\frac{x^n}{x^{n+1}}\right|=\lim_{n\to\infty}\left|\frac2x\right|=\frac2{|x|}\lim_{n\to\infty}1=\frac2{|x|}$

Convergence requires that this limit be smaller than 1, so this happens when $\dfrac2{|x|}2$ . (This agrees with what you already know about geometric series; $\dfrac{2^n}{x^n}=\left(\dfrac2x\right)^n$ only converges if $\left|\dfrac2x\right|$ .)

The ratio test also works with the second problem.

$\displaystyle\lim_{n\to\infty}\left|\frac{\frac{\sin^{n+1}x}{3^{n+1}}}{\frac{\sin^nx}{3^n}}\right|=\lim_{n\to\infty}\left|\frac{\sin x}3\right|=\dfrac{|\sin x|}3$
$\implies |\sin x|$

But recall that for all

, you have $|\sin x|\le1$ , so the above condition is always met. This means the series converges everywhere.

These first two are also good candidates for using the root test.

For the last two, try the integral test.

$\dfrac1{n(\ln n)^p}$ will be positive for all

so long as $n\ge3$ ( $\ln2$ is negative, and so the first term in the series will be negative for any odd

; but we can just omit this term from the series because any one term is still a finite number). So, we compare to the integral,

$\displaystyle\int_3^\infty\frac{\mathrm dx}{x(\ln x)^p}\le\sum_{n\ge3}\frac1{n(\ln n)^p}$

Let $y=\ln x$ , so that $\mathrm dy=\dfrac{\mathrm dx}x$ . Then

$\displaystyle\int_3^\infty\frac{\mathrm dx}{x(\ln x)^p}=\int_{\ln3}^\infty\frac{\mathrm dy}{y^p}$
$=\displaystyle\lim_{t\to\infty}\int_{\ln3}^t\frac{\mathrm dy}{y^p}$
$=\displaystyle\lim_{t\to\infty}\frac{y^{1-p}}{1-p}\bigg|_{y=\ln3}^{y=t}$
$=\displaystyle\lim_{t\to\infty}\frac{t^{1-p}}{1-p}-\frac{(\ln3)^{1-p}}{1-p}$

Note that we immediately know $p\neq1$ . You have

$\displaystyle\lim_{t\to\infty}\frac{t^{1-p}}{1-p}=\begin{cases}\infty&\text{for }1-p\ge0\\0&\text{for }1-p$

so the integral converges only for $1-p\le0\implies p1$ .

For the last one, omit the first two terms ( $n=1\implies\dfrac{\ln n}{n^p}=0$ and $n=2\implies\dfrac{\ln n}{n^p}$ ). The remaining sequence is positive and decreasing, which follows from the fact that $\ln n$ . Then compare to the definite integral,

$\displaystyle\int_3^\infty\frac{\ln x}{x^p}\,\mathrm dx\le\sum_{n\ge3}\frac{\ln n}{n^p}$

Integrate by parts, setting $u=\ln x$ and $\mathrm dv=x^{-p}\,\mathrm dx$ , so that $\mathrm du=\dfrac{\mathrm dx}x$ and $v=\dfrac{x^{1-p}}{1-p}$ . Then

$\displaystyle\int_3^\infty\frac{\ln x}{x^p}\,\mathrm dx=\frac{x^{1-p}\ln x}{1-p}\bigg|_{x=3}^{x\to\infty}-\frac1{1-p}\int_3^\infty x^{2-p}\,\mathrm dx$
$=\displaystyle\lim_{t\to\infty}\frac{t^{1-p}\ln t}{1-p}-\frac{(\ln3)^{1-p}\ln(\ln3)}{1-p}-\frac{x^{1-p}}{(1-p)^2}\bigg|_{x=\ln3}^{x\to\infty}$
$=\displaystyle\lim_{t\to\infty}\left(\frac{t^{1-p}\ln t}{1-p}-\frac{t^{1-p}}{(1-p)^2}\right)+\frac{(\ln3)^{1-p}}{(1-p)^2}-\frac{(\ln3)^{1-p}\ln(\ln3)}{1-p}$
$=\displaystyle\frac1{1-p}\lim_{t\to\infty}\left(\frac{t^{1-p}\ln t}{1-p}-\frac{t^{1-p}}{(1-p)^2}\right)+\frac{(\ln3)^{1-p}}{(1-p)^2}-\frac{(\ln3)^{1-p}\ln(\ln3)}{1-p}$

Again, we can't have p=1

. The integral will converge if the limit converges to a finite number.

$\displaystyle\lim_{t\to\infty}\left(\frac{t^{1-p}\ln t}{1-p}-\frac{t^{1-p}}{(1-p)^2}\right)=\frac1{(1-p)^2}\lim_{t\to\infty}t^{1-p}((p-1)\ln t-1)$

When

, the limit diverges to $\infty$ . When

, evaluating the limit directly gives an indeterminate form $0\times\infty$ . We rewrite this as

$\displaystyle\lim_{t\to\infty}\frac{(p-1)\ln t-1}{t^{p-1}}$

and using the fact that $\ln t$ for large enough

whenever

, it follows that the limit exists by the squeeze theorem. So L'Hopital's rule applies here, and the limit is equal to

$\displaystyle\lim_{t\to\infty}\frac{(p-1)\ln t-1}{t^{p-1}}=\lim_{t\to\infty}\frac{\frac{p-1}t}{(p-1)t^{p-2}}=\lim_{t\to\infty}t^{p-1}=0$

All this to say the last series converges for

Answer from: Quest

solution:

→c= a b , that is product of a and b is possible only when when a is m × n matrix and b is n × p matrix.

that is, number of columns of a= number of rows of b.

so, size of matrix , c= a b is , if a = (m × n)matrix

b= (n × p)matrix

then a b = [ (m × n)matrix] ×[ (n × p)matrix ]

→→c= a b= [m × p ] matrix

→→size of product a b will be equal to = number of rows of a × number of columns of b

Answer from: Quest

eve and marks ratios are equivalent

step-by-step explanation:

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Convergence/divergence of the sums of series!

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