determine whether the series is absolutely convergent, conditionally convergent or divergent. If it converges, estimate the remainder r10.

Question

sum of {n=1} to infinity of (-1)^n ((n^2)/sqrt(n^5+n+1))

NEED ASAP!! please provide step by step solution and answers.

Thank you!

Answer 1

First we deal with absolute convergence, so we drop the (-1)^n, and every term of n^2/sqrt(n^5 + n + 1) is non-negative

Consider (n^5 + n + 1) < (n^5 + n^5 + n^5) = 3n^5

Then, sqrt(n^5 + n + 1) < sqrt(3n^5)

Then, 1/sqrt(n^5 +n + 1) > 1/sqrt(3n^5)

And, n^2/sqrt(n^5 + n + 1) > n^2/sqrt(3n^5) = n^2/(sqrt(3)*n^2.5) = 1/(sqrt(3)*n^(0.5))

Now, sum of {n=1 to inf} of 1/(sqrt(3)*n^(0.5)) diverges by the p-series, since p = 0.5 < 1.

Further, every term of n^2/sqrt(n^5 + n + 1) is greater than the corresponding term of 1/(sqrt(3)*n^(0.5))

Therefore, sum of {n=1 to inf} of n^2/sqrt(n^5 + n + 1) is absolutely divergent by comparison test (that is, the given sum is not absolutely convergent).

Now we consider whether the given sum is conditionally convergent. We observe that each term in n^2/sqrt(n^5 + n + 1) is larger than the next term and the limit of n^2/sqrt(n^5 + n + 1) = 0. Thus, the given sum will converge by the alternating series test; the given sum is conditionally convergent.

Answer 2

The series in which a_n = 1/n^p converges for p>1 and diverges for p less than or equal to 1.

For n>1, n² > n+n^-3+n^-4

n>sqrt(n+n^-3+n^-4) => 1/n<1/sqrt(n+n^-3+n^-4)

Hence the series cannot converge absolutely, but the alternating series converges by Leibnitz's theorem, i.e. this is a so-called conditionally convergent series.

You will probably need to do some reading/studying to understand this answer.

If not in your textbook, get M. Spivak's textbook Calculus (4th edition) and look in the chapter on infinite series.