Test the series for convergence or divergence. ∑from n=1 to ∞ ((n^3 -1)/(n^4+1))

I do the alternating seriestest. It shows that bn >0, (bn+1)<bn and the limit is 0. So, it is convergent but the correct answer is divergent. Why?

why alternating series? All the terms are positive.

lim(n->∞) a_(n+1)/(a_n) = n^3/n^4 = 1/n

The harmonic series diverges

Test the series for convergence or divergence.∑from n=1 to ∞ ((e^n) /n^2)

n = 1

To test the series for convergence or divergence, we can use the Limit Comparison Test.

Step 1: Determine the terms of the series.
The terms of the given series are ((n^3 - 1)/(n^4 + 1)).

Step 2: Find a comparison series.
We need to find a comparison series whose convergence or divergence is already known.

Since we have polynomials of degree 3 and 4, it is useful to compare the given series with a p-series. A p-series has the form ∑(1/n^p), where p is a positive constant.

In this case, let's take p = 4. So, the p-series is ∑(1/n^4).

Step 3: Take the limit of the ratio.
We need to calculate the limit as n approaches infinity of the ratio of the terms of the given series and the p-series.

Let's denote the given series as A(n) = ((n^3 - 1)/(n^4 + 1)) and the p-series as B(n) = 1/n^4.

We can write the limit comparison as:

L = lim(n -> ∞) (A(n) / B(n))

L = lim(n -> ∞) (((n^3 - 1)/(n^4 + 1)) / (1/n^4))

L = lim(n -> ∞) ((n^3 - 1)/(n^4 + 1)) * (n^4/1)

Simplifying further:

L = lim(n -> ∞) (n^7 - n^4) / (n^4 + 1)

Step 4: Evaluate the limit.
To evaluate the limit, we divide the leading terms in the numerator and denominator by the highest power of n.

L = lim(n -> ∞) (n^3 - 1/n^4) / (1 + 1/n^4)

As n approaches infinity, the constants (1/n^4 and 1) become negligible. Therefore, they do not contribute significantly to the limit.

L = lim(n -> ∞) (n^3 / n^4)

L = lim(n -> ∞) (1 / n)

Step 5: Analyze the limit.
Since the limit as n approaches infinity of (1/n) is equal to zero, we can conclude that the limit comparison is zero.

Step 6: Make the conclusion.
According to the Limit Comparison Test, if the limit comparison is a finite positive number, the series and the comparison series either both converge or both diverge.

Since the limit comparison is zero, it means that the given series ((n^3 - 1)/(n^4 + 1)) and the comparison series (∑(1/n^4)) converge or diverge together.

The series ∑((n^3 - 1)/(n^4 + 1)) is a convergent series.

Therefore, the given series converges.