show if it is divergent or convergent......?

Question

show if it is divergent or convergent......

summation of 1/sqrt(n^6 + 4n^2 + 3) from 1 to infinity.....

please guys!!!!!

i dont know how to figure this out....

Answer 1

Easily done. Note that since n will be positive for all of the terms in the series, we will have

n^6 + 4n^2 + 3 > n^6 for all terms, and so

1/ sqrt (n^6 + 4n^2 + 3) < 1 / sqrt (n^6)

= 1 / n^3 for all n.

We can see that the summation from one to infinity of 1 / n^3 is convergent using the integral test, the terms of our given series are all positive, and since they are dominated, term by term, by the terms of a convergent series, our given series is convergent.

Some people, at a moment like this, will urge you to do a partial fractions decomposition. That will seldom be anything other than a time wasting distraction, since the terms don't tend to telescope, and you tend to be left with something of the form infinity minus infinity. What are you going to do with that? Best to keep things simple, when you can.

Answer 2

For every n =1, 2,3...sqrt(n^6 + 4n^2 + 3) > sqrt(n^6) = n^3 (all the terms in the sqrt are positive). Therefore, also for every n,

0 <1/sqrt(n^6 + 4n^2 + 3) < 1/n^3 (1)

We know the series Sum(1/n^p), n=1,2,3.... converges if p>1 and diverges if p <=1. Therefore, Sum (1/n^3) converges. Considering (1), we conclude, by comparisson, that Sum (n =1, oo) 1/sqrt(n^6 + 4n^2 + 3) converges. And we can affirm that Sum (n =1, oo) 1/sqrt(n^6 + 4n^2 + 3) < Sum(n=1, oo) (1/n^3).

Hope this helps.

Answer 3

Use a comparison test.

1/√(n^6 + 4n^2 + 3) < 1/√n^6 = 1/n³
which converges since the exponent is > 1.

Therefore 1/√(n^6 + 4n^2 + 3) converges.

Answer 4

ARGH AP/GP!

IM TERRIFIED OF THIS TOPIC