anonymous
  • anonymous
Consider the power series (n^10(x+7)^n)/((5^n)(n^(32/3))) need to find the interval of convergence and radius of convergence, please help
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
freckles
  • freckles
Have you tried applying the ratio test?
anonymous
  • anonymous
i have but am not fully understand it yet
freckles
  • freckles
this is what we want: \[|\frac{a_{n+1}}{a_n}|<1 \] well first we can simplify the left hand expression a bit \[|a_{n+1} \cdot \frac{1}{a_n}|=|\frac{(n+1)^{10} (x+7)^{n+1}}{5^{n+1}( n+1)^\frac{32}{3}} \cdot \frac{5^n n^\frac{32}{3}}{n^{10}}(x+7)^n | \]

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

freckles
  • freckles
for example \[\frac{5^n}{5^{n+1}}=\frac{1}{5} \\ \frac{(x+7)^{n+1}}{(x+7)^n}=x+7\] and yes that (x+7)^n was suppose to be in the bottom
freckles
  • freckles
\[|a_{n+1} \cdot \frac{1}{a_n}|=|\frac{(n+1)^{10}(x+7)^{n+1}}{5^{n+1}(n+1)^\frac{32}{3}} \cdot \frac{5^n n^\frac{32}{3}}{n^{10} (x+7)^n}| \\ |\frac{a_{n+1}}{a_n}|=\frac{1}{5}|x+7| \cdot |\frac{(n+1)^{10} n^\frac{32}{3}}{(n+1)^\frac{32}{3}n^{10}}|\] and we have n goes to infinity
freckles
  • freckles
so you need to find the limit of that thing as n goes to infinity and we want that to be less than 1
anonymous
  • anonymous
okay thanks, can u show me the answer or show me who to get there?
anonymous
  • anonymous
to find the interval of convergence and radius of convergence for it
freckles
  • freckles
\[\frac{1}{5} |x+7| | \lim_{n \rightarrow \infty } \frac{(n+1)^{10}n^\frac{32}{3}}{(n+1)^{\frac{32}{3}}n^{10}}|<1\] well you first need to evaluate that limit I was talking about then you can solve for x
anonymous
  • anonymous
so just look at1/5(x+7) how do u eval limit?
freckles
  • freckles
are you saying you don't know how to evaluate this limit? \[\lim_{n \rightarrow \infty} \frac{(n+1)^{10} n^{\frac{32}{3}}}{(n+1)^\frac{32}{3}n^{10}}\]
freckles
  • freckles
\[\lim_{n \rightarrow \infty} \frac{n^{\frac{32}{3}-10}}{(n+1)^{\frac{32}{3}-10}} \\ =\lim_{n \rightarrow \infty} (\frac{n}{n+1})^\frac{2}{3} \\ = (\lim_{n \rightarrow \infty}\frac{n}{n+1})^\frac{2}{3}\] can you evaluate the limit now ?
anonymous
  • anonymous
is it 1?
freckles
  • freckles
yes the inside thing goes to 1 and 1^(2/3) is 1
freckles
  • freckles
\[\frac{1}{5} |x+7| | \lim_{n \rightarrow \infty } \frac{(n+1)^{10}n^\frac{32}{3}}{(n+1)^{\frac{32}{3}}n^{10}}|<1 \\ \frac{1}{5} |x+7| |1|<1 \\ \frac{1}{5} |x+7|<1\]
anonymous
  • anonymous
is this for R or X?
anonymous
  • anonymous
is x then 35?
freckles
  • freckles
what does that mean ? Solve for x to find the interval of convergence you might want to consider testing the endpoints since at L=1 the ratio test is inconclusive I don't see how you got x=35 when we had an inequality to solve.
freckles
  • freckles
recall |f|