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anonymous
 one year ago
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
anonymous
 one year ago
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

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0your help is greatly appreciated

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1lmLtfy \(\huge \frac{10^n}{(n+1)6^{(2n+1)}} \) is that it?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0are u able to help, i really need it

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1so in ratio test we compare successive terms, right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes i for come reason could get a value for L and then can't tell if it diverges or converges

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0can u help me find L

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i know it converges but can't find the value, would love help! to find L

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1haven't tried this in ages so here goes: \( \huge \frac{\frac{10^{n+1}}{(n+2)6^{(2n+3)}}}{\frac{10^n}{(n+1)6^{(2n+1)}}} \) \( \huge = \frac{\frac{10^{n}10^{1}}{(n+2)6^{2n}6^{3}}}{\frac{10^n}{(n+1)6^{2n}6^{1}}} \) \( \huge = \frac{\frac{10}{(n+2)6^{3}}}{\frac{1}{(n+1)6^{1}}} \) \( \huge = \frac{10}{(n+2)6^{3}} \times \frac{(n+1)6}{1} = \frac{10}{6^2}\frac {n+1}{n+2}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0does that mean L =10/36???

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0can someone explain what L is?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0tangential 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 $$\frac1{1x}=\sum_{n=0}^\infty x^n\\\implies \log(1x)=\sum_{n=0}^\infty \frac1{n+1}x^{n+1}$$ so we have that $$S=\frac35\log\left(1\frac5{18}\right)=\frac35\log\left(\frac{13}{18}\right)=\frac35\left(\log13\log18\right)$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0or i guess that's actually \(\log(1x)=\sum_{n=0}^\infty\frac1{n+1}x^{n+1}\) so we actually have that $$S=\frac35\log\left(1\frac5{18}\right)=\frac35\log\left(\frac{18}{13}\right)=\frac35(\log18\log13)$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so L=3/4(log18log13) can anyone verify this is correct? thanks again or your help, as i need help finding the correct L value

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no, that's not \(L\)  that's the actual value of the series

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0for \(L\) follow what @IrishBoy123 says

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you get $$L=\lim_{n\to\infty}\frac{10}{36}\cdot\frac{n+1}{n+2}=\frac5{18}\lim_{n\to\infty}\frac{1+1/n}{1+2/n}=\frac5{18}<1$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh so 5/18 is L then?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1@nick1234567 L is 10 / 36 i think you saw that

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yep, \(L=5/18=10/36\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but the series itself sums to \(\frac35(\log2+2\log3\log13)\approx0.195253\)

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.0\[ \begin{align*} f(n)&=\frac{10^n}{(n+1)6^{2n+1}}\\ &=\frac{10^n}{{6(n+1)36^n}}\\ &=\frac{1}{6(n+1)} \left( \frac{5}{18} \right)^{n}\\ \end{align*} \] \[ \begin{align*} \lim_{n\to\infty}\left\frac{f(n+1)}{f(n)}\right&=\lim_{n\to\infty}\left\frac{1}{6(n+2)} \left( \frac{5}{18} \right)^{n+1}6(n+1)\left(\frac{18}{5}\right)^n\right\\ &=\lim_{n\to\infty}\left\frac{n+1}{n+2}\frac{5}{18}\right\\ &=\frac{5}{18} \end{align*} \] My math is definitely rusty.
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