Estimating Error for an Infinite Series (Mclaurin)

[kq6up]

Homework Statement

Problem # 30 in Ch1 Section 16 in Mary L. Boas' Math Methods in the Physical Sciences

It is clear that you (or your computer) can’t find the sum of an infinite series
just by adding up the terms one by one. For example, to get \zeta (1.1)=\sum _{ n=1 }^{ \infty }{ \frac { 1 }{ { n }^{ 1.1 } } } (see Problem 15.22) with error < 0.005 takes about 10^{33} terms.
To see a simple alternative (for a series of positive decreasing terms) look at
Figures 6.1 and 6.2. Show that when you have summed N terms, the sum RN
of the rest of the series is between { I }_{ N }=\int _{ N }^{ \infty }{ { a }_{ n } } dn\quad and\quad { I }_{ N+1 }=\int _{ N+1 }^{ \infty }{ { a }_{ n } } dn

Homework Equations

They are above.

The Attempt at a Solution

I am not sure how I am supposed to integrate an a_{n} since it contains n!.

Thanks,
Chris Maness

 

[Mark44]

Homework Statement

Problem # 30 in Ch1 Section 16 in Mary L. Boas' Math Methods in the Physical Sciences

It is clear that you (or your computer) can’t find the sum of an infinite series
just by adding up the terms one by one. For example, to get \zeta (1.1)=\sum _{ n=1 }^{ \infty }{ \frac { 1 }{ { n }^{ 1.1 } } } (see Problem 15.22) with error < 0.005 takes about 10^{33} terms.
To see a simple alternative (for a series of positive decreasing terms) look at
Figures 6.1 and 6.2. Show that when you have summed N terms, the sum RN
of the rest of the series is between { I }_{ N }=\int _{ N }^{ \infty }{ { a }_{ n } } dn\quad and\quad { I }_{ N+1 }=\int _{ N+1 }^{ \infty }{ { a }_{ n } } dn

Homework Equations

They are above.

The Attempt at a Solution

I am not sure how I am supposed to integrate an a_{n} since it contains n!.

Thanks,
Chris Maness

You haven't told us what a_n is.

 

[kq6up]

a_n is the coefficient for the Mclaurin series. a_n=f(x)^n/n!

Chris

 

[pasmith]

Homework Statement

Problem # 30 in Ch1 Section 16 in Mary L. Boas' Math Methods in the Physical Sciences

It is clear that you (or your computer) can’t find the sum of an infinite series
just by adding up the terms one by one. For example, to get \zeta (1.1)=\sum _{ n=1 }^{ \infty }{ \frac { 1 }{ { n }^{ 1.1 } } } (see Problem 15.22) with error < 0.005 takes about 10^{33} terms.
To see a simple alternative (for a series of positive decreasing terms) look at
Figures 6.1 and 6.2. Show that when you have summed N terms, the sum RN
of the rest of the series is between { I }_{ N }=\int _{ N }^{ \infty }{ { a }_{ n } } dn\quad and\quad { I }_{ N+1 }=\int _{ N+1 }^{ \infty }{ { a }_{ n } } dn

Homework Equations

They are above.

The Attempt at a Solution

I am not sure how I am supposed to integrate an a_{n} since it contains n!.

Thanks,
Chris Maness

The more serious problem would be attempting to integrate f^{(n)}(0) with respect to n.

The wording "To see a simple alternative (for a series of positive decreasing terms)" suggests that a_n is an arbitrary sequence of positive decreasing terms which decreases fast enough for \sum a_n to converge, and not the Mclaurin series of an arbitrary function (which may contain negative terms).

In any event, you are not asked to do the actual integration, but to explain why, if a_n is a decreasing positive sequence and f : \mathbb{R} \to \mathbb{R} is a decreasing continuous positive function such that f(n) = a_n for all n, it is the case that
<br /> \int_N^{\infty} f(x)\,dx \leq \sum_{N+1}^\infty a_n \leq \int_{N+1}^\infty f(x)\,dx. <br />
I suggest you look at the figures you are invited to look at, and see if that suggests a reason.