Rudin's explanation of how rapid the series 1/(n) converges

In summary: Multiplying this by 1/(n+1)! gives 1/n!, which is equal to the right hand side of the equation. Therefore, the equality holds.
  • #1
DeadOriginal
274
2
In Rudin's Principle's of Mathematical Analysis, Rudin days that we can estimate how fast the series [itex]\sum\frac{1}{n!}[/itex] converges by the following:
Put
$$
s_{n}=\sum_{k=0}^{n}\frac{1}{k!}.
$$
Then
$$
e-s_{n}
=\frac{1}{(n+1)!}+\frac{1}{(n+2)!}+\frac{1}{(n+3)!}+\cdots<\frac{1}{(n+1)!}\left(1+\frac{1}{n+1}+\frac{1}{(n+1)^{2}}+\cdots\right)=\frac{1}{n!n}
$$
so that
$$
0<e-s_{n}<\frac{1}{n!n}.
$$
The part that bothers me is
$$
\frac{1}{(n+1)!}\left(1+\frac{1}{n+1}+\frac{1}{(n+1)^{2}}+\cdots\right)=\frac{1}{n!n}.
$$
Using Maple I was able to see that
$$
\frac{1}{(n+1)!}\sum_{k=0}^{\infty}\frac{1}{(n+1)^{k}}=\frac{1}{n!n}
$$
but what if I did not have access to anything like Maple or Mathematica. How would I be able to figure out that the equality holds?
 
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  • #2
That sum is a geometric series - do you know what to do with those?
 
  • #3
Oh. HAHAHAHA. Wow. Okay. I see it now. Thanks.
 
  • #4
DeadOriginal said:
In Rudin's Principle's of Mathematical Analysis, Rudin days that we can estimate how fast the series [itex]\sum\frac{1}{n!}[/itex] converges by the following:
Put
$$
s_{n}=\sum_{k=0}^{n}\frac{1}{k!}.
$$
Then
$$
e-s_{n}
=\frac{1}{(n+1)!}+\frac{1}{(n+2)!}+\frac{1}{(n+3)!}+\cdots<\frac{1}{(n+1)!}\left(1+\frac{1}{n+1}+\frac{1}{(n+1)^{2}}+\cdots\right)=\frac{1}{n!n}
$$
so that
$$
0<e-s_{n}<\frac{1}{n!n}.
$$
The part that bothers me is
$$
\frac{1}{(n+1)!}\left(1+\frac{1}{n+1}+\frac{1}{(n+1)^{2}}+\cdots\right)=\frac{1}{n!n}.
$$
Using Maple I was able to see that
$$
\frac{1}{(n+1)!}\sum_{k=0}^{\infty}\frac{1}{(n+1)^{k}}=\frac{1}{n!n}
$$
but what if I did not have access to anything like Maple or Mathematica. How would I be able to figure out that the equality holds?

The sum is simply a geometric series with r = 1/(n+1), so the sum = 1/(1-r) = (n+1)/n.
 
  • #5


In Rudin's explanation, the main idea is that the terms in the series decrease very rapidly, so we can estimate the remainder by looking at the first few terms that we have not yet summed. However, it is not necessary to use Maple or Mathematica to understand why the equality holds.

Firstly, we can notice that the series on the right-hand side is a geometric series with a common ratio of $\frac{1}{n+1}$. We can use the formula for the sum of a geometric series to rewrite it as $\frac{1}{(n+1)!}\cdot \frac{1}{1-\frac{1}{n+1}}$. Simplifying this gives us $\frac{1}{n!n}$.

Now, on the left-hand side, we have a sum of terms where each term is $\frac{1}{(n+1)!}$ multiplied by a power of $\frac{1}{n+1}$. We can rewrite this as $\frac{1}{(n+1)!}\cdot \frac{1}{(n+1)^k}$, where $k$ is the power of $\frac{1}{n+1}$. So we can see that the sum on the left-hand side is equivalent to $\frac{1}{(n+1)!}\cdot \sum_{k=0}^{\infty}\frac{1}{(n+1)^k}$. And as we saw before, this is equal to $\frac{1}{n!n}$.

In summary, the equality holds because the series on the left-hand side can be rewritten as a geometric series and simplified to match the series on the right-hand side. This is a fundamental property of geometric series and does not require the use of any computer software to understand.
 

1. What is Rudin's explanation of the rapid convergence of the series 1/(n)?

Rudin's explanation is based on the comparison test, which states that if a series is absolutely convergent, then any series with smaller terms will also be absolutely convergent. Since the series 1/(n) is absolutely convergent, any series with terms smaller than 1/(n) will also be absolutely convergent, making the convergence rapid.

2. Why is the series 1/(n) considered to have rapid convergence?

The series 1/(n) is considered to have rapid convergence because it converges faster than many other series, meaning that the terms approach 0 quickly. This is due to the fact that the terms decrease in size as n increases, making the series easier to sum and resulting in a faster convergence rate.

3. How does Rudin's explanation of the rapid convergence of 1/(n) relate to the convergence rate of other series?

Rudin's explanation of the rapid convergence of 1/(n) shows that the series has a faster convergence rate than many other series. This means that the terms approach 0 at a faster rate, resulting in a quicker convergence and making the series easier to sum.

4. Can Rudin's explanation be applied to other series with similar terms?

Yes, Rudin's explanation can be applied to other series with similar terms, as long as those series are absolutely convergent. This is because the comparison test can be used to show that any series with smaller terms than 1/(n) will also be absolutely convergent, resulting in a rapid convergence rate.

5. Are there any limitations to Rudin's explanation of the rapid convergence of 1/(n)?

One limitation of Rudin's explanation is that it only applies to series with terms that decrease in size as n increases. If the terms do not decrease in size, the comparison test cannot be used and the convergence rate may not be as rapid. Additionally, the comparison test can only be used if the series being compared is absolutely convergent.

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