Question

how do I find were P converges in the series of:
the sum from n=1 to infinity of (7n^7/(4n^5+2))^p

Answer 1

the degree of the denominator has to be more than one larger than the degree of the numerator

Answer 2

for example 2 would work because then the degree of the denominator would be 10 and that is 3 larger than 7
of course you can pick a smaller p

Answer 3

ca you explain a little more? i haven't fully grasped the concept

Answer 4

\[\sum\limits_{n=1}^{\infty}\left(\dfrac{7n^7}{4n^5+2}\right)^p\]

like this ?

Answer 5

yes

Answer 6

i need to find all values were p converges

Answer 7

so that exponent p is not just for the denominator term
its for the entire fraction is it ?

Answer 8

yes it is for the entire fraction

Answer 9

\[\sum\limits_{n=1}^{\infty}\left(\dfrac{7n^7}{4n^5+2}\right)^p\]
The series will diverge if the numerator degree is \(k\ge1\). If we set \(p=1\), then we can see that the series diverges because we have a "cumulative" degree of \(k=2\). (In other words, dividing a 7th degree polynomial by a 5th degree polynomial yields a 2nd degree polynomial.)

You want to find \(p\) such that you can essentially normalize this degree so that it always is less than 1.

Hint: Think rational numbers. You can ignore most of the algebra involved with expansions. What value of \(p\) times 2 gives you something less than 1?

Answer 10

you may use p-series test here
just notice that "highest exponent" of numerator = 7p
"highest exponent" of denominator = 5p

Answer 11

\[\sum\limits_{n=1}^{\infty}\left(\dfrac{7n^7}{4n^5+2}\right)^p \lt C\sum\limits_{n=1}^{\infty}\dfrac{1}{n^{-2p}}\]

appealing to p-series test gives you the same answer as siths

Answer 12

oh okay I see