The two problems require to approximate the series to four decimal points. both are assumed to be convergent
1) sin2n/n5 from 1 to inifinity
I have no idea what to do because Its not an alternating series and therefore I cant use the alternating series est thm and I dont have advanced enough learning to know how to take the integral for the intergral approx test.

2)(-1)n-1(n)/(n4+1) from 1 to infinity

"Both are assumed to be convergent" - then why do you think you would need to use the Alternating Series test or Integral test to check its convergence?

If you are wanting to find a series to represent \displaystyle \begin{align*} \frac{\sin^2{(n)}}{n^5} \end{align*}, I'd write \displaystyle \begin{align*} \sin^2{(n)} = \frac{1}{2} - \frac{1}{2}\cos{(2n)} \end{align*} and use the Maclaurin series for \displaystyle \begin{align*} \cos{(2n)} \end{align*}.

sorry i didnt specify not the alternating series test for convergence, i'm talking about the alternating series estimation therom (ie. |S-Sn| < bn+1)
and the intergral estimation sum:
integral(n+1 to infinity) f(x) < Rn < integral (n to infinity) f(x)

where S= Value the series converges to
Sn= nth Partial Sum
Rn= Remainder or 'error'

Hi,
In your statement of an integral test, who is function f? Answer: f is a decreasing non-negative function on say $[1,\infty)$ with $f(n)=a_n$ and $\int_1^\infty f(x)\,dx$ is finite. Here also $R_n=\sum_{k=n+1}^\infty a_n$.

$$b_n={sin^2(n)\over n^5}\text{. Then }R_n=|\sum_{k=n+1}^\infty b_n|\leq\sum_{k=n+1}^\infty {1\over k^5}$$

So let $f(x)={1\over x^5}$. Use this to find an error bound for $\sum_{k=n+1}^\infty {1\over k^5}$ and hence for $R_n$

Your second series is an alternating series with the terms decreasing.

Originally Posted by johng
Hi,
In your statement of an integral test, who is function f? Answer: f is a decreasing non-negative function on say $[1,\infty)$ with $f(n)=a_n$ and $\int_1^\infty f(x)\,dx$ is finite. Here also $R_n=\sum_{k=n+1}^\infty a_n$.

$$b_n={sin^2(n)\over n^5}\text{. Then }R_n=|\sum_{k=n+1}^\infty b_n|\leq\sum_{k=n+1}^\infty {1\over k^5}$$

So let $f(x)={1\over x^5}$. Use this to find an error bound for $\sum_{k=n+1}^\infty {1\over k^5}$ and hence for $R_n$

Your second series is an alternating series with the terms decreasing.
I thought you were unable to apply the alternating series estimation thm to a none alternating series though??

Originally Posted by smoez
I thought you were unable to apply the alternating series estimation thm to a none alternating series though??
$\sum_{n=1}^\infty \frac{(-1)^{n-1}n}{n^4+1}$ is an alternating series - whose terms decrease in absolute value. But you are correct - you can't apply the alternating series estimation theorem to a non-alternating series.

- Hollywood

Originally Posted by hollywood
$\sum_{n=1}^\infty \frac{(-1)^{n-1}n}{n^4+1}$ is an alternating series - whose terms decrease in absolute value. But you are correct - you can't apply the alternating series estimation theorem to a non-alternating series.

- Hollywood
I'm talking about the sum: sin^2(n)/n^5.

Yeah - that one's not an alternating series (in fact, all terms are positive), so you need to use johng's method.

- Hollywood

Originally Posted by johng
Hi,
In your statement of an integral test, who is function f? Answer: f is a decreasing non-negative function on say $[1,\infty)$ with $f(n)=a_n$ and $\int_1^\infty f(x)\,dx$ is finite. Here also $R_n=\sum_{k=n+1}^\infty a_n$.

$$b_n={sin^2(n)\over n^5}\text{. Then }R_n=|\sum_{k=n+1}^\infty b_n|\leq\sum_{k=n+1}^\infty {1\over k^5}$$

So let $f(x)={1\over x^5}$. Use this to find an error bound for $\sum_{k=n+1}^\infty {1\over k^5}$ and hence for $R_n$

Your second series is an alternating series with the terms decreasing.
I'm sorry but im still very confused by this solution could you please walk me through what ideas you used it seems like a mix of the alternating series therom and comparison test but i cant be sure

He's bounding the error, which is exactly $\left| \sum_{k=n+1}^\infty b_n \right|$ by the easier-to-analyze series $\sum_{k=n+1}^\infty \frac{1}{k^5}$. That's similar to the comparison test; he's looking at the actual sum instead of just testing for convergence.
$\sum_{k=n+1}^\infty \frac{1}{k^5} \le \int_n^\infty \frac{1}{x^5}\,dx$.