Does the following series converges or diverges?

Question

∞

∑ (n2-3n+5)/(n3+4n-1)

n=1

Answer 1

This series can be shown to diverge by using the comparison test.

 

Since the +5 and -1 terms don't do much to the overall series, the series will behave in a similar way as

 

(n²-3n)/(n³+4n)

 

We can reduce this a bit to make things easier to see by factoring out one of the n terms

 

(n²-3n)/(n³+4n) = n(n-3)/n(n²+4) = (n-3)/(n²-4)

 

Again, the -3 and -4 terms don't do much, ignore them.

 

The series will behave in a similar way as 

 

n/n² = 1/n

 

And since we know that 1/n is harmonic, we know that the entire series will diverge!

 

 

Keep in mind that this will not always be conclusive for a convergent series. Just because a smaller series converges, does not mean that a larger one will not diverge.

 

Answer 2

Here's a more formal proof that the series diverges:

 

∑(n²-3n+5)/(n³+4n-1) > ∑(n²-3n)/(n³+4n-1) > ∑(n²-3n)/(n³+5n) = ∑(n-3)/(n²+5) = ∑(n)/(n²+5) - ∑(3)/(n²+5)

  > ∑(n)/(n²+5n) - ∑(3)/(6n²) = ∑(1)/(n+5) - (1/2)*∑(1/n²) > ∑(1/6n) - (1/2)*∑(1/n²) = (1/6)*∑(1/n) - (1/2)*∑(1/n²)

 

*Note, that the first summation diverges while the second summation converges, giving a net diverging series. Thus, by comparision test, since the initial series is greater than a diverging series, the initial series diverges. Each inequality assumes the terms in series contain values when n is greater than 1.

Answer 3

We use the Divergence test here. If the limit of the function as n approaches infinity is NOT equal to zero, then the function diverges.

 

Let's take a look at your function now. You have a squared term in the numerator and a cubic term in the denominator, which results in a limit of 1/x (which is a very special case).

 

It turns out that 1/x is divergent, so the series diverges as well.

 

My thanks to Darek for the correction!