Examining both of the series below. Which would you choose as the correct response from the solutions below?

Question

oo
∑ (n^-1)/(n^2 + 1)
n=1

oo
∑ n^4/(n^5 + 1)
n=1

possible responses....

a. Only the first series converges.

b. Only the second series converges.

c. Both series converge.

d. Both series diverge.

Answer 1

Only the first one converges. A

It's not difficult to see that the first one converges because is reduces to 1 / (n^3 + n). Using loose terminology, the high power of n in the denominator assures that the final term goes to zero.

It's the second one that's tricky.
Let's rewrite it as follows:
(n^4 + 1/n - 1/n ) / (n^5 + 1)
= (n^4 + 1/n ) / (n^5 + 1) - (1/n ) / (n^5 + 1)
= 1/n - 1/(n^6 + n)
This is a sum of two series. The second term converges, but we know for a fact that 1/n does not converge. So this series does not converge.

NOTE: ∑ 1/n = - ln(1-x) as x goes to 1.

Answer 2

I would say (a).

The first series converges by the p-test. After pulling down the x^(-1) from the numerator, you have a denominator with p=3>1.

The second series diverges by a comparison test with the harmonic series. Term-by-term it is smaller than the harmonic series which diverges, thus it diverges.

Answer 3

a)
The first is of the same class as 1/n^3 convergent and the 2nd is of the same class as 1/n divergent

Answer 4

a.
coz
the first series is equivalenet to the sum of 1/n^3
the second is equivalent to the sum of 1/n
(Reimann series)
if we took the sum of 1/n^p
if p<=1 the series div.
if p>1 the series conv.