Summation n = 1 to infinity

(2n^4 - n^3 + 7n^2 - 1) / (3n^5 + 1)

prove

Tried limit comparison, comparison, divergence test, ratio test. Can't figure it out :/

This post was edited by OnlyD3 on May 19 2013 04:41pm

First you should try to analyze the series to see what you expect and that often gives some insight into how to do the problem

We have a ratio of polynomials; the top is of the order of n^4 and the bottom of n^5.
But 4^2/n^5 = 1/n and we know that the series summation 1/n diverges (harmonic series). So we expect our series to behave like the harmonic series and diverge; further this suggests we try to compare our series to the harmonic series to prove divergence.
(another way to see the above that might be simplier to you is to note that for very large n, the top is essentially only n^4; and again, when n is very large, the n^5 on the bottom is much greater than the n^4-n^3...-1, so the bottom is approx n^5 for large n. Thus our series is approx n^2/n^3 which = 1/n and summation 1/n diverges)


The inequalitites I used are standard tricks: to make a polynomial larger, replace each term with the highest power; to make it smaller, try half the highest power term)

You could also do this problem with the limit comparison test- using 1/n to compare



Have fun with series..

Thanks. So I wouldn't use bn = 1/n^5 to compare because the numerator has a polynomial of order > 1 ?

i wouldn't compare it to that for that reason to be honest.

I immediately just saw a comparison of polynomials, so all you need to consider is the dominant terms.
In saying that, i haven't done serious series qns for a couple years, so i am a tad rusty

I see. So why not use 1/n^5 to compare?

(2n^4 - n^3 + 7n^2 - 1) / (3n^5 + 1) ~ 2n^4 / 3n^4 (that is 2/3n) when n -> infinity. Since 2/3n is a positive term with a divergent serie, (2n^4 - n^3 + 7n^2 - 1) / (3n^5 + 1) has a divergent serie too.

It will lead you nowhere if you try to compare to 1/n^5.

But you could manually prove that it is greater than, for say, 1/2n, with n high enough :

(2n^4 - n^3 + 7n^2 - 1) / (3n^5 + 1) > 1/2n for every n > N

( N = 0 is working )

and since 1/2n has a divergent serie, yours is divergent too.

diverges

can be shown simply by finding a f which is smaller than your function numerator

find g larger than your function denom.

clearly f/g < y

show that f and g diverge

...

profit?

because that's not what basic form it factors into

factor out n^4 from the top, n^5 from bottom, and cancel out