Series question (from a calc 2 class)

[geliot]

Hi,
So I'm looking at a review sheet for a test i have tomorrow, and most of it makes sense but there's one question on series that I don't see how to solve/answer.

"Let S sub n denote the nth partial sum of the series ?(1/(n+1)^2) from n=1 to infinity.
Explain why the following inequality is valid (with a graph/sentences).
?(1/(n+1)^2)< S sub n + [integral from n to infinity of] 1/(x+1)^2dx.

Then use the inequality (above) to provide an estimate of the series that is accurate to within .01."

I graphed it something like this
http://i188.photobucket.com/albums/z108 ... /graph.jpg (its pretty rough, sorry!), and I guess that means that S sub n will always be greater than the sum of the series up to n (the little bit above the curved line, right?)? Is this true though? I don't really see why...

When it comes to using the inequality to provide an estimate, I'm at a complete loss. I know how to solve for geometric series, and for alternating series, but I'm at a loss for this one - i definitely don't see how to use the inequality; does anybody have any ideas?

Thank you so much,
Eliot.

 

[galactus]

\(\displaystyle \sum_{n=1}^{\infty}\frac{1}{(n+1)^{2}}\)

This is the rather famous Euler series.

\(\displaystyle \frac{1}{2^{2}}+\frac{1}{3^{2}}+\frac{1}{4^{2}}+................\)

The sum of the reciprocals of the squares minus the 1.

Thus, the sum is \(\displaystyle \frac{{\pi}^{2}}{6}-1\)

But:

\(\displaystyle s_{n}+\int_{n+1}^{\infty}f(x)dx<S<s_{n}+\int_{n}^{\infty}f(x)dx\)

This provides an upper and lower bound on the sum S of the series.

Where S is the sum of the infinite series and \(\displaystyle s_{n}\) is the partial sum.

\(\displaystyle \int_{n+1}^{\infty}\frac{1}{(x+1)^{2}}dx=\frac{1}{n+2}\)

\(\displaystyle \int_{n}^{\infty}\frac{1}{(x+1)^{2}}dx=\frac{1}{n+1}\)

If \(\displaystyle S=\sum_{k=1}^{\infty}u_{k}\) and \(\displaystyle s_{n}=\sum_{k=1}^{n}u_{k}\),

then \(\displaystyle S-s_{n}=\sum_{k=n+1}^{\infty}u_{k}\)

Interpret \(\displaystyle u_{k}, \;\ k=n+1, \;\ n+2, ......,\) as the area of inscribed or circumscribed rectangles

with height \(\displaystyle u_{k}\) and base of length 1 for the curve \(\displaystyle y=f(x)\) to obtain the result.

Here is an example with **\(\displaystyle \sum_{k=1}^{\infty}\frac{1}{k^{3}}\).

Let use \(\displaystyle s_{10}\) as the partial sum of the series and use it to get an upper and lower bound on S.

\(\displaystyle s_{10}\approx 1.1975\)

Let \(\displaystyle f(x)=\frac{1}{x^{3}}\), then \(\displaystyle \int_{10}^{\infty}\frac{1}{x^{3}}dx=\frac{1}{200}=.005\)

and \(\displaystyle \int_{11}^{\infty}\frac{1}{x^{3}}dx=\frac{1}{242}\approx .0041\)

so \(\displaystyle 1.2016<S<1.2026\)

If we rewrite the inequality as \(\displaystyle \int_{n+1}^{\infty}f(x)dx<S-s_{n}<\int_{n}^{\infty}f(x)dx\)

then \(\displaystyle \int_{n}^{\infty}\frac{1}{x^{3}}dx=\frac{1}{2n^{2}}<\frac{1}{1000}, \;\ n>\sqrt{500}\approx 22.4; \;\ n=23\)

This ensures the error is within 1/1000.

**If one studies the Riemann Zeta function, we find this is \(\displaystyle \zeta(3)\)