Maclaurin series and Riemann Sum

Question

Find the Maclaurin series using e¹ for the lim n->∞ nsin(2πen!):

The Maclaurin series for e×=1+x+x²/2!+x³/3!+...=∑x^k/k! as k goes from 0 to ∞. Time=Distance/Rate.

Is it e¹/nsin(2πen!)=1/nsin(2πn!)? Or is the Maclaurin series for e× adjusted to remove the first two terms in the series, i.e., 1+x?

 

A>-1 and b>-1, use Riemann sum to compute lim n->∞ n^b-a (1^a+2^a+3^a...n^a/1^b+2^b+3^b...n^b). 

Answer 1

Hi Anthony!

Great questions! These problems use two important calculus concepts: Maclaurin series and Riemann sums. Let's tackle them separately.

Problem 1: Maclaurin Series

We want to evaluate:

[

\lim_{n\to\infty}n\sin(2\pi e,n!)

]

The key observation is that (e) is very close to a rational number whose denominator is (n!).

Step 1: Start with the Maclaurin series for (e^x).

We know:

[

e^x=\sum_{k=0}^{\infty}\frac{x^k}{k!}

]

Setting (x=1):

[

e=1+1+\frac{1}{2!}+\frac{1}{3!}+\cdots

]

Multiply both sides by (n!):

[

en!=n!\sum_{k=0}^{\infty}\frac{1}{k!}

]

Separate this into two parts:

[

en!=\sum_{k=0}^{n}\frac{n!}{k!}

+\sum_{k=n+1}^{\infty}\frac{n!}{k!}

]

The first sum is an integer because (n!/k!) is an integer whenever (k\le n).

Call that integer (M_n).

Then:

[

en!=M_n+R_n

]

where:

[

R_n=\sum_{k=n+1}^{\infty}\frac{n!}{k!}

]

Step 2: Simplify the sine expression.

Substitute into the original expression:

[

\sin(2\pi en!)

=\sin(2\pi M_n+2\pi R_n)

]

Because sine has period (2\pi), and (M_n) is an integer:

[

\sin(2\pi en!)=\sin(2\pi R_n)

]

Therefore:

[

n\sin(2\pi en!)=n\sin(2\pi R_n)

]

Step 3: Determine the behavior of the remainder.

The remainder is:

[

R_n=\frac{1}{n+1}

+\frac{1}{(n+1)(n+2)}+\cdots

]

As (n) becomes large:

[

R_n\sim\frac{1}{n+1}

]

Therefore:

[

nR_n\to1

]

Since (R_n\to0), we can use the familiar limit:

[

\lim_{x\to0}\frac{\sin x}{x}=1

]

Rewrite:

2\pi(nR_n)

\frac{\sin(2\pi R_n)}{2\pi R_n}

]

Taking the limit:

[

=2\pi(1)(1)

]

Final Answer:

[

\boxed{2\pi}

]

Notice that we do not simply remove the first two terms of the Maclaurin series. Instead, we separate the expansion into an integer portion and a remainder. The integer portion disappears inside the sine function because of periodicity.

Problem 2: Riemann Sums

We want to evaluate:

[

\lim_{n\to\infty}

n^{b-a}

\left(

\frac{1^a+2^a+\cdots+n^a}

{1^b+2^b+\cdots+n^b}

\right)

]

where (a>-1) and (b>-1).

Step 1: Rewrite the expression.

Using summation notation:

[

L=\lim_{n\to\infty}

n^{b-a}

\frac{\sum_{k=1}^{n}k^a}

{\sum_{k=1}^{n}k^b}

]

We want to rewrite the sums so they resemble Riemann sums.

Divide the numerator and denominator by the appropriate powers of (n):

[

L=

\lim_{n\to\infty}

\frac{

\frac{1}{n}\sum_{k=1}^{n}(k/n)^a

}{

\frac{1}{n}\sum_{k=1}^{n}(k/n)^b

}

]

Step 2: Recognize the Riemann sums.

Recall that:

\int_0^1 f(x),dx

]

For the numerator:

\int_0^1x^a,dx

]

Since (a>-1):

[

\int_0^1x^a,dx=\frac{1}{a+1}

]

Similarly, for the denominator:

[

\int_0^1x^b,dx=\frac{1}{b+1}

]

The conditions (a>-1) and (b>-1) ensure that both integrals converge, even when the exponents are negative.

Step 3: Evaluate the limit.

Substitute our results:

[

L=

\frac{\frac{1}{a+1}}

{\frac{1}{b+1}}

]

Simplifying:

[

\boxed{L=\frac{b+1}{a+1}}

]

Final Answers

Problem 1:

[

\boxed{2\pi}

]

Problem 2:

[

\boxed{\frac{b+1}{a+1}}

]

Key takeaway: In the first problem, the important idea is combining a Maclaurin expansion with the periodicity of sine. In the second, the trick is recognizing that sums of powers can be rewritten as Riemann sums and evaluated using definite integrals.

Hope this helps clarify both approaches!

Michael