## 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

1. anonymous

2. IrishBoy123

lmLtfy $$\huge \frac{10^n}{(n+1)6^{(2n+1)}}$$ is that it?

3. anonymous

yes

4. anonymous

are u able to help, i really need it

5. IrishBoy123

so in ratio test we compare successive terms, right?

6. anonymous

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

7. anonymous

can u help me find L

8. anonymous

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

9. IrishBoy123

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

does that mean L =10/36???

11. anonymous

can someone explain what L is?

12. anonymous

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

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

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

no, that's not $$L$$ -- that's the actual value of the series

16. anonymous

for $$L$$ follow what @IrishBoy123 says

17. anonymous

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

oh so 5/18 is L then?

19. IrishBoy123

@nick1234567 L is 10 / 36 i think you saw that

20. anonymous

yep, $$L=5/18=10/36$$

21. anonymous

thank U!!!!

22. anonymous

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

23. thomas5267

\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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