Question

Ratio test on:\[\frac{ 10n^3 2^{n+4}}{\pi^n}\]
I went \[\frac{ 10 }{ 1 } \times \frac{2^{n+4} }{2^n } \times \frac{ \pi^n }{ \pi^{n+1} }\]

and got 320/pi, which is obviously bigger than 1 and divergent by the ratio test. However, my book tells me that I can use the ratio test to get a convergent number. How do I do that?

Answer 1

Ratio test on:\[\frac{ 10n^3 2^{n+4}}{\pi^n}\]
I went \[\frac{ 10 }{ 1 } \times \frac{2^{n+4} }{2^n } \times \frac{ \pi^n }{ \pi^{n+1} }\]

and got 320/pi, which is obviously bigger than 1 and divergent by the ratio test. However, my book tells me that I can use the ratio test to get a covergent number. How do I do that?

Answer 2

Hmmm...
\[\Large \frac{10(n+1)^32^{n+5}}{\pi^{n+1}}\times\frac{\pi^n}{10n^32^{n+4}}\]

Answer 3

Is this what you did?

Answer 4

>.>
Anyway... things happen...
\[\Large =\left(\frac{n+1}{n}\right)^3\times\frac{2}{\pi}\]

Answer 5

And you'll find that when n gets really really really big, the whole expression just becomes essentially
\[\huge \frac{2}{\pi}<1\]

Answer 6

Thanks for showing me my mistake!

Answer 7

No problem :)