Question

the series ((10^n)/((n+1)6^(2n+1)) what does L = in the ratio test and does it converge or diverge by the test

Answer 1

your help is greatly appreciated

Answer 2

lmLtfy

\(\huge \frac{10^n}{(n+1)6^{(2n+1)}} \)

is that it?

Answer 3

yes

Answer 4

are u able to help, i really need it

Answer 5

so in ratio test we compare successive terms, right?

Answer 6

yes i for come reason could get a value for L and then can't tell if it diverges or converges

Answer 7

can u help me find L

Answer 8

i know it converges but can't find the value, would love help! to find L

Answer 9

haven't tried this in ages so here goes:

\( \huge \frac{\frac{10^{n+1}}{(n+2)6^{(2n+3)}}}{\frac{10^n}{(n+1)6^{(2n+1)}}} \)

\( \huge = \frac{\frac{10^{n}10^{1}}{(n+2)6^{2n}6^{3}}}{\frac{10^n}{(n+1)6^{2n}6^{1}}} \)

\( \huge = \frac{\frac{10}{(n+2)6^{3}}}{\frac{1}{(n+1)6^{1}}} \)

\( \huge = \frac{10}{(n+2)6^{3}} \times \frac{(n+1)6}{1} = \frac{10}{6^2}\frac {n+1}{n+2}\)

Answer 10

does that mean L =10/36???

Answer 11

can someone explain what L is?

Answer 12

tangential but here is its closed form:

$$S=\sum_{n=0}^\infty\frac{10^n}{(n+1)6^{2n+1}}=\frac16\sum_{n=0}^\infty\frac1{n+1}\left(\frac{10}{36}\right)^n=\frac35\sum_{n=0}^\infty\frac1{n+1}\left(\frac5{18}\right)^{n+1}$$

now consider $$\frac1{1-x}=\sum_{n=0}^\infty x^n\\\implies \log(1-x)=\sum_{n=0}^\infty \frac1{n+1}x^{n+1}$$ so we have that $$S=\frac35\log\left(1-\frac5{18}\right)=\frac35\log\left(\frac{13}{18}\right)=\frac35\left(\log13-\log18\right)$$

Answer 13

or i guess that's actually \(\log(1-x)=-\sum_{n=0}^\infty\frac1{n+1}x^{n+1}\) so we actually have that $$S=-\frac35\log\left(1-\frac5{18}\right)=\frac35\log\left(\frac{18}{13}\right)=\frac35(\log18-\log13)$$

Answer 14

so L=3/4(log18-log13) can anyone verify this is correct? thanks again or your help, as i need help finding the correct L value

Answer 15

no, that's not \(L\) -- that's the actual value of the series

Answer 16

for \(L\) follow what @IrishBoy123 says

Answer 17

you get $$L=\lim_{n\to\infty}\frac{10}{36}\cdot\frac{n+1}{n+2}=\frac5{18}\lim_{n\to\infty}\frac{1+1/n}{1+2/n}=\frac5{18}<1$$

Answer 18

oh so 5/18 is L then?

Answer 19

@nick1234567

L is 10 / 36

i think you saw that

Answer 20

yep, \(L=5/18=10/36\)

Answer 21

thank U!!!!

Answer 22

but the series itself sums to \(\frac35(\log2+2\log3-\log13)\approx0.195253\)

Answer 23

\[
\begin{align*}
f(n)&=\frac{10^n}{(n+1)6^{2n+1}}\\
&=\frac{10^n}{{6(n+1)36^n}}\\
&=\frac{1}{6(n+1)} \left( \frac{5}{18} \right)^{n}\\
\end{align*}
\]
\[
\begin{align*}
\lim_{n\to\infty}\left|\frac{f(n+1)}{f(n)}\right|&=\lim_{n\to\infty}\left|\frac{1}{6(n+2)} \left( \frac{5}{18} \right)^{n+1}6(n+1)\left(\frac{18}{5}\right)^n\right|\\
&=\lim_{n\to\infty}\left|\frac{n+1}{n+2}\frac{5}{18}\right|\\
&=\frac{5}{18}
\end{align*}
\]
My math is definitely rusty.