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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

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  1. anonymous
    • one year ago
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    your help is greatly appreciated

  2. IrishBoy123
    • one year ago
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    lmLtfy \(\huge \frac{10^n}{(n+1)6^{(2n+1)}} \) is that it?

  3. anonymous
    • one year ago
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    yes

  4. anonymous
    • one year ago
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    are u able to help, i really need it

  5. IrishBoy123
    • one year ago
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    so in ratio test we compare successive terms, right?

  6. anonymous
    • one year ago
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    yes i for come reason could get a value for L and then can't tell if it diverges or converges

  7. anonymous
    • one year ago
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    can u help me find L

  8. anonymous
    • one year ago
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    i know it converges but can't find the value, would love help! to find L

  9. IrishBoy123
    • one year ago
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    haven't tried this in ages so here goes: \( \huge \frac{\frac{10^{n+1}}{(n+2)6^{(2n+3)}}}{\frac{10^n}{(n+1)6^{(2n+1)}}} \) \( \huge = \frac{\frac{10^{n}10^{1}}{(n+2)6^{2n}6^{3}}}{\frac{10^n}{(n+1)6^{2n}6^{1}}} \) \( \huge = \frac{\frac{10}{(n+2)6^{3}}}{\frac{1}{(n+1)6^{1}}} \) \( \huge = \frac{10}{(n+2)6^{3}} \times \frac{(n+1)6}{1} = \frac{10}{6^2}\frac {n+1}{n+2}\)

  10. anonymous
    • one year ago
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    does that mean L =10/36???

  11. anonymous
    • one year ago
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    can someone explain what L is?

  12. anonymous
    • one year ago
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    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 $$S=\frac35\log\left(1-\frac5{18}\right)=\frac35\log\left(\frac{13}{18}\right)=\frac35\left(\log13-\log18\right)$$

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

  14. anonymous
    • one year ago
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    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

  15. anonymous
    • one year ago
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    no, that's not \(L\) -- that's the actual value of the series

  16. anonymous
    • one year ago
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    for \(L\) follow what @IrishBoy123 says

  17. anonymous
    • one year ago
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    you get $$L=\lim_{n\to\infty}\frac{10}{36}\cdot\frac{n+1}{n+2}=\frac5{18}\lim_{n\to\infty}\frac{1+1/n}{1+2/n}=\frac5{18}<1$$

  18. anonymous
    • one year ago
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    oh so 5/18 is L then?

  19. IrishBoy123
    • one year ago
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    @nick1234567 L is 10 / 36 i think you saw that

  20. anonymous
    • one year ago
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    yep, \(L=5/18=10/36\)

  21. anonymous
    • one year ago
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    thank U!!!!

  22. anonymous
    • one year ago
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    but the series itself sums to \(\frac35(\log2+2\log3-\log13)\approx0.195253\)

  23. thomas5267
    • one year ago
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    \[ \begin{align*} f(n)&=\frac{10^n}{(n+1)6^{2n+1}}\\ &=\frac{10^n}{{6(n+1)36^n}}\\ &=\frac{1}{6(n+1)} \left( \frac{5}{18} \right)^{n}\\ \end{align*} \] \[ \begin{align*} \lim_{n\to\infty}\left|\frac{f(n+1)}{f(n)}\right|&=\lim_{n\to\infty}\left|\frac{1}{6(n+2)} \left( \frac{5}{18} \right)^{n+1}6(n+1)\left(\frac{18}{5}\right)^n\right|\\ &=\lim_{n\to\infty}\left|\frac{n+1}{n+2}\frac{5}{18}\right|\\ &=\frac{5}{18} \end{align*} \] My math is definitely rusty.

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