Question

WILL GIVE MEDAL
Use the ratio test to find the radius of convergence of the power series
x + 4 x^2 + 9 x^3 + 16 x^4 + 25 x^5 + \cdots

Answer 1

what would the equation that you do the ratio test by

Answer 2

@binarymimic please help

Answer 3

i know \[a _{n}=C _{n}(x-a)^{n}\]

Answer 4

where \[(x-a)^{n}=(x)^{n+1}\]

Answer 5

sorry, was not at computer. can you rewrite this in summation notation? you will find it looks like

\[\sum_{k = 1}^{\infty} k ^{2}x ^{k}\]

from here take limit of next term divided by previous term.. i.e.,

\[\lim_{k \rightarrow \infty} \frac{ (k+1)^{2}x ^{k+1} }{ k ^{2}x ^{k} }\]

Answer 6

should be in absolute value, but it simplifies greatly to just |x|. the ratio test will tell us exactly for what values this series converges

Answer 7

wouldnt it start at k=0 tho

Answer 8

i think your first term is just x

Answer 9

because the first term is x

Answer 10

k^2 x^k, k = 0 gives 0 as your first term

Answer 11

ohhh

Answer 12

are there like general steps to finding out what the function will look like or do you just have to figure it out differently every time.

Answer 13

you will have to determine the pattern of the series and write it in a notation such that, given an index, n, you can determine the next term in the sequence (n+1). see http://www.sosmath.com/calculus/radcon/radcon02/radcon02.html

i don't know of any general formula, each series is different

Answer 14

but for this particular one -- the series ratio simplifies to just |x|. that is, as k goes to infinite, the ratio of the next term to the previous term approaches |x|. the ratio test states that, if the ratio of the next term to previous term is < 1, then the ratio converges. in the case where the ratio is |x|, the series converges whenever |x| < 1