Can we somehow modify the Lagrange form to get a tighter bound? (Curious)

[mathmari]

Hey!

I am looking at the following:

1. Show that $\displaystyle{\text{exp}(1)=\sum_{k=0}^{\infty}\frac{1}{k!}=e}$ with $\displaystyle{e:=\lim_{n\rightarrow \infty}\left (1+\frac{1}{n}\right )^n}$.
   
   Hint: Use the binomial theorem and compare with the partial sum $s_n$ of the series $\sum_{k=0}^{\infty}\frac{1}{k!}$.
2. Calculate $s_n$ and $\left (1+\frac{1}{n}\right )^n$ for $n=10$ and $n=100$ with exactly $12$ decimal places and compare with $e$.
   
3. Show that the differenz of the $n$-th parial sum of the series $\sum_{k=0}^{\infty}\frac{1}{k!}$ can be estimated to $e$ by $0<e-s_n<\frac{1}{n!n}$.

I have done the following:

1. We have the partial sum $$s_n=\sum_{k=0}^n\frac{1}{k!}$$
   
   By the binomial theorem we have that $$\left (1+\frac{1}{n}\right )^n=\sum_{k=0}^n\binom{n}{k}\frac{1}{n^k}$$
   
   We have that \begin{align*}\binom{n}{k}\frac{1}{n^k}&=\frac{n!}{k!\cdot (n-k)!}\cdot \frac{1}{n^k}\\ & =\frac{(n-k+1) \cdot \ldots \cdot (n-1) \cdot n}{k!}\cdot \frac{1}{n^k}\\ & =\frac{1}{k!}\cdot \frac{(n-k+1)\cdot \ldots \cdot (n-1) \cdot n}{n^k}\\ & =\frac{1}{k!}\cdot \frac{n-k+1}{n}\cdot \ldots \cdot \frac{n-1}{n}\cdot \frac{n}{n}\\ & =\frac{1}{k!}\cdot \left (1-\frac{k-1}{n}\right )\cdot \ldots \cdot \left (1-\frac{1}{n}\right )\cdot 1\\ & \leq \frac{1}{k!}\end{align*}We want to find also an upper bound.
   
   By the binomial theorem we have that $$\left (1+\frac{1}{n+1}\right )^{n+1}=\sum_{k=0}^{n+1}\binom{n+1}{k}\frac{1}{(n+1)^k}$$
   
   \begin{align*}\binom{n+1}{k}\frac{1}{(n+1)^k}&=\frac{(n+1)!}{k!\cdot ([n+1]-k)!}\cdot \frac{1}{(n+1)^k} \\ & =\frac{([n+1]-k+1)\cdot \ldots \cdot n\cdot (n+1)}{k!}\cdot \frac{1}{(n+1)^k} \\ & \geq \frac{([n+1]-k+1)\cdot \ldots \cdot ([n+1]-k+1)\cdot ([n+1]-k+1)}{k!}\cdot \frac{1}{(n+1)^k} \\ & =\frac{1}{k!}\cdot \frac{([n+1]-k+1)\cdot \ldots \cdot ([n+1]-k+1)\cdot ([n+1]-k+1)}{(n+1)^k}\\ & =\frac{1}{k!}\cdot \frac{[n+1]-k+1}{n+1}\cdot \ldots \cdot \frac{[n+1]-k+1}{n+1}\cdot \frac{[n+1]-k+1}{n+1} \end{align*}
   
   How can we continue? (Wondering) So, we will get \begin{align*}&\binom{n}{k}\frac{1}{n^k}\leq \frac{1}{k!}\leq \binom{n+1}{k}\frac{1}{(n+1)^k} \\ & \Rightarrow \sum_{k=0}^n\binom{n}{k}\frac{1}{n^k}\leq \sum_{k=0}^n\frac{1}{k!}\leq \sum_{k=0}^n \binom{n+1}{k}\frac{1}{(n+1)^k}\leq \sum_{k=0}^{n+1} \binom{n+1}{k}\frac{1}{(n+1)^k} \\ & \Rightarrow \left (1+\frac{1}{n}\right )^n \leq s_n \leq \left (1+\frac{1}{n+1}\right )^{n+1} \\ & \Rightarrow \lim_{n\rightarrow \infty}\left (1+\frac{1}{n}\right )^n \leq \lim_{n\rightarrow \infty}s_n \leq \lim_{n\rightarrow \infty} \left (1+\frac{1}{n+1}\right )^{n+1} \\ & \Rightarrow e \leq \sum_{k=0}^{\infty}\frac{1}{k!} \leq e \\ & \Rightarrow \sum_{k=0}^{\infty}\frac{1}{k!}=e\end{align*}
   
   Is everything correct? (Wondering)
2. Do we calculate $s_n$ and $\left (1+\frac{1}{n}\right )^n$ for $n=10$ and $n=100$ with exactly $12$ decimal places with the calculator, or is it meant to do something else here? (Wondering)
   
3. We have that
   $$e-s_n=\sum_{k=0}^{\infty}\frac{1}{k!}-\sum_{k=0}^n\frac{1}{k!}=\sum_{k=n+1}^{\infty}\frac{1}{k!}$$
   We have that $\sum_{k=n+1}^{\infty}\frac{1}{k!}$ is popsitive, sinve every term is positiv. So, $0<e-s_n$.
   
   Could you give me a hint how we could get the upper bound? (Wondering)

 

[DrWahoo]

Could always use generating functions for the binomial.

Hey!

I am looking at the following:

1. Show that $\displaystyle{\text{exp}(1)=\sum_{k=0}^{\infty}\frac{1}{k!}=e}$ with $\displaystyle{e:=\lim_{n\rightarrow \infty}\left (1+\frac{1}{n}\right )^n}$.
   
   Hint: Use the binomial theorem and compare with the partial sum $s_n$ of the series $\sum_{k=0}^{\infty}\frac{1}{k!}$.
2. Calculate $s_n$ and $\left (1+\frac{1}{n}\right )^n$ for $n=10$ and $n=100$ with exactly $12$ decimal places and compare with $e$.
   
3. Show that the differenz of the $n$-th parial sum of the series $\sum_{k=0}^{\infty}\frac{1}{k!}$ can be estimated to $e$ by $0<e-s_n<\frac{1}{n!n}$.

I have done the following:

1. We have the partial sum $$s_n=\sum_{k=0}^n\frac{1}{k!}$$
   
   By the binomial theorem we have that $$\left (1+\frac{1}{n}\right )^n=\sum_{k=0}^n\binom{n}{k}\frac{1}{n^k}$$
   
   We have that \begin{align*}\binom{n}{k}\frac{1}{n^k}&=\frac{n!}{k!\cdot (n-k)!}\cdot \frac{1}{n^k}\\ & =\frac{(n-k+1) \cdot \ldots \cdot (n-1) \cdot n}{k!}\cdot \frac{1}{n^k}\\ & =\frac{1}{k!}\cdot \frac{(n-k+1)\cdot \ldots \cdot (n-1) \cdot n}{n^k}\\ & =\frac{1}{k!}\cdot \frac{n-k+1}{n}\cdot \ldots \cdot \frac{n-1}{n}\cdot \frac{n}{n}\\ & =\frac{1}{k!}\cdot \left (1-\frac{k-1}{n}\right )\cdot \ldots \cdot \left (1-\frac{1}{n}\right )\cdot 1\\ & \leq \frac{1}{k!}\end{align*}We want to find also an upper bound.
   
   By the binomial theorem we have that $$\left (1+\frac{1}{n+1}\right )^{n+1}=\sum_{k=0}^{n+1}\binom{n+1}{k}\frac{1}{(n+1)^k}$$
   
   \begin{align*}\binom{n+1}{k}\frac{1}{(n+1)^k}&=\frac{(n+1)!}{k!\cdot ([n+1]-k)!}\cdot \frac{1}{(n+1)^k} \\ & =\frac{([n+1]-k+1)\cdot \ldots \cdot n\cdot (n+1)}{k!}\cdot \frac{1}{(n+1)^k} \\ & \geq \frac{([n+1]-k+1)\cdot \ldots \cdot ([n+1]-k+1)\cdot ([n+1]-k+1)}{k!}\cdot \frac{1}{(n+1)^k} \\ & =\frac{1}{k!}\cdot \frac{([n+1]-k+1)\cdot \ldots \cdot ([n+1]-k+1)\cdot ([n+1]-k+1)}{(n+1)^k}\\ & =\frac{1}{k!}\cdot \frac{[n+1]-k+1}{n+1}\cdot \ldots \cdot \frac{[n+1]-k+1}{n+1}\cdot \frac{[n+1]-k+1}{n+1} \end{align*}
   
   How can we continue? (Wondering) So, we will get \begin{align*}&\binom{n}{k}\frac{1}{n^k}\leq \frac{1}{k!}\leq \binom{n+1}{k}\frac{1}{(n+1)^k} \\ & \Rightarrow \sum_{k=0}^n\binom{n}{k}\frac{1}{n^k}\leq \sum_{k=0}^n\frac{1}{k!}\leq \sum_{k=0}^n \binom{n+1}{k}\frac{1}{(n+1)^k}\leq \sum_{k=0}^{n+1} \binom{n+1}{k}\frac{1}{(n+1)^k} \\ & \Rightarrow \left (1+\frac{1}{n}\right )^n \leq s_n \leq \left (1+\frac{1}{n+1}\right )^{n+1} \\ & \Rightarrow \lim_{n\rightarrow \infty}\left (1+\frac{1}{n}\right )^n \leq \lim_{n\rightarrow \infty}s_n \leq \lim_{n\rightarrow \infty} \left (1+\frac{1}{n+1}\right )^{n+1} \\ & \Rightarrow e \leq \sum_{k=0}^{\infty}\frac{1}{k!} \leq e \\ & \Rightarrow \sum_{k=0}^{\infty}\frac{1}{k!}=e\end{align*}
   
   Is everything correct? (Wondering)
2. Do we calculate $s_n$ and $\left (1+\frac{1}{n}\right )^n$ for $n=10$ and $n=100$ with exactly $12$ decimal places with the calculator, or is it meant to do something else here? (Wondering)
   
3. We have that
   $$e-s_n=\sum_{k=0}^{\infty}\frac{1}{k!}-\sum_{k=0}^n\frac{1}{k!}=\sum_{k=n+1}^{\infty}\frac{1}{k!}$$
   We have that $\sum_{k=n+1}^{\infty}\frac{1}{k!}$ is popsitive, sinve every term is positiv. So, $0<e-s_n$.
   
   Could you give me a hint how we could get the upper bound? (Wondering)

 

[I like Serena]

1. We want to find also an upper bound.

By the binomial theorem we have that $$\left (1+\frac{1}{n+1}\right )^{n+1}=\sum_{k=0}^{n+1}\binom{n+1}{k}\frac{1}{(n+1)^k}$$

Hey mathmari! (Smile)

I don't think this is going to work, because if we inspect a couple of values with for instance Excel, we'll see that this won't give us an upper bound. That's because $s_n$ converges much faster than $(1+\frac 1n)^n$.

2. Do we calculate $s_n$ and $\left (1+\frac{1}{n}\right )^n$ for $n=10$ and $n=100$ with exactly $12$ decimal places with the calculator, or is it meant to do something else here?

I think a calculator is okay, although we need to be careful to avoid serious rounding errors.
With a regular calculator, $1/k!$ will become so small with respect to $s_n$ that adding it will not actually add it.
Btw, I think finding these numbers will show that your approach to find the upper bound won't work. (Thinking)

3. We have that
$$e-s_n=\sum_{k=0}^{\infty}\frac{1}{k!}-\sum_{k=0}^n\frac{1}{k!}=\sum_{k=n+1}^{\infty}\frac{1}{k!}$$
We have that $\sum_{k=n+1}^{\infty}\frac{1}{k!}$ is popsitive, sinve every term is positiv. So, $0<e-s_n$.

Could you give me a hint how we could get the upper bound?

I think we need to figure out to solve (1) first, although we can probably use the given upper bound as a hint.

To be honest, I haven't figured out yet how to find the upper bound. (Worried)
I did find a possible hint on wiki:

The number $e$ is the unique real number such that
$$\left(1+\frac{1}{x}\right)^x < e < \left(1+\frac{1}{x}\right)^{x+1}$$
for all positive $x$.
​

(Thinking)

 

[DrWahoo]

Gould's Home Page

Vol 1 - 8 all contain good info on this type of stuff.

 

[mathmari]

I did find a possible hint on wiki:

The number $e$ is the unique real number such that
$$\left(1+\frac{1}{x}\right)^x < e < \left(1+\frac{1}{x}\right)^{x+1}$$
for all positive $x$.
​

(Thinking)

We have the following:
\begin{align*}\left (1+\frac{1}{n}\right )^n&=\sum_{k=0}^n\binom{n}{k}\frac{1}{n^k}\\ & = \sum_{k=0}^n\frac{n!}{k!\cdot (n-k)!}\cdot \frac{1}{n^k} \\ & = \sum_{k=0}^n\frac{(n-k+1)\cdot (n-k+2)\cdot \ldots \cdot (n-1)\cdot n}{k!}\cdot \frac{1}{n^k} \\ &= \sum_{k=0}^n\frac{n-k+1}{n}\cdot \frac{n-k+2}{n}\cdot \ldots \cdot \frac{n-1}{n}\cdot \frac{n}{n}\cdot \frac{1}{k!} \\ & = \sum_{k=0}^n\left (1-\frac{k-1}{n}\right )\cdot \left (1-\frac{k-2}{n}\right )\cdot \ldots \cdot \left (1-\frac{1}{n}\right )\cdot 1\cdot \frac{1}{k!} \\ & \leq \sum_{k=0}^n \frac{1}{k!}\end{align*}

We also have the following:
\begin{align*}\left (1+\frac{1}{n}\right )^{n+1}&=\sum_{k=0}^{n+1}\binom{n+1}{k}\cdot \frac{1}{n^k} \\ & = 1+\sum_{k=1}^{n+1}\binom{n+1}{k}\cdot\frac{1}{n^k} \\ & \overset{ m:=k-1 }{ = } 1+\sum_{m=0}^{n}\binom{n+1}{m+1}\cdot\frac{1}{n^{m+1}} \\ & \overset{ k:=m }{ = } 1+\sum_{k=0}^{n}\binom{n+1}{k+1}\cdot\frac{1}{n^{k+1}} \\ & = 1+\sum_{k=0}^{n}\frac{(n+1)!}{(k+1)!\cdot ([n+1]-[k+1])!}\cdot\frac{1}{n^{k+1}} \\ & = 1+\sum_{k=0}^{n}\frac{(n+1)!}{(k+1)!\cdot (n-k)!}\cdot\frac{1}{n^{k+1}} \\ & = 1+\sum_{k=0}^{n}\frac{(n-k+1)\cdot \ldots n\cdot (n+1)}{(k+1)!}\cdot\frac{1}{n^{k+1}} \\ & \geq 1+\sum_{k=0}^{n}\frac{(n-k+1)\cdot \ldots (n-k+1)\cdot (n-k+1)}{(k+1)!}\cdot\frac{1}{n^{k+1}} \\ & = 1+\sum_{k=0}^{n}\frac{(n-k+1)^{k+1}}{(k+1)!}\cdot\frac{1}{n^{k+1}} \\ & = 1+\sum_{k=0}^{n}\frac{(n-k+1)^{k+1}}{n^{k+1}}\cdot\frac{1}{(k+1)!} \\ & = 1+\sum_{k=0}^{n}\left (\frac{n-k+1}{n}\right )^{k+1}\cdot\frac{1}{(k+1)!} \\ & = 1+\sum_{k=0}^{n}\left (1-\frac{k-1}{n}\right )^{k+1}\cdot\frac{1}{(k+1)!} \\ & \overset{\text{Bernoulli}}{ \geq } 1+\sum_{k=0}^{n}\left [1-\frac{k-1}{n}\cdot (k+1)\right ]\cdot\frac{1}{(k+1)!} \\ & = 1+\sum_{k=0}^{n}\frac{1}{(k+1)!}-\sum_{k=0}^{n}\frac{(k-1)\cdot (k+1)}{n}\cdot\frac{1}{(k+1)!} \\ & = 1+\sum_{k=0}^{n}\frac{1}{(k+1)!}-\frac{1}{n}\cdot \sum_{k=0}^{n}\frac{(k-1)}{k!} \\ & \overset{ m:=k+1}{ = } 1+\sum_{m=1}^{n+1}\frac{1}{m!}-\frac{1}{n}\cdot \sum_{k=0}^{n}\frac{(k-1)}{k!} \\ & \overset{ k:=m}{ = } 1+\sum_{k=1}^{n+1}\frac{1}{k!}-\frac{1}{n}\cdot \sum_{k=0}^{n}\frac{(k-1)}{k!} \\ & = \frac{1}{0!}+\sum_{k=1}^{n}\frac{1}{k!}+\frac{1}{(n+1)!}-\frac{1}{n}\cdot \sum_{k=0}^{n}\frac{(k-1)}{k!} \\ & = \sum_{k=0}^{n}\frac{1}{k!}+\frac{1}{(n+1)!}-\frac{1}{n}\cdot \sum_{k=0}^{n}\frac{(k-1)}{k!} \\ & \geq \sum_{k=0}^{n}\frac{1}{k!}-\frac{1}{n}\cdot \sum_{k=0}^{n}\frac{(k-1)}{k!} \\ & = \sum_{k=0}^{n}\frac{1}{k!}-\frac{1}{n}\cdot \left [-1+0+\sum_{k=2}^{n}\frac{(k-1)}{k!}\right ] \\ & = \sum_{k=0}^{n}\frac{1}{k!}+\frac{1}{n}-\frac{1}{n}\cdot \sum_{k=2}^{n}\left (\frac{k}{k!}-\frac{1}{k!}\right ) \\ & = \sum_{k=0}^{n}\frac{1}{k!}+\frac{1}{n}-\frac{1}{n}\cdot \sum_{k=2}^{n}\left (\frac{1}{(k-1)!}-\frac{1}{k!}\right ) \\ & \overset{\text{telescoping series}}{ = } \sum_{k=0}^{n}\frac{1}{k!}+\frac{1}{n}-\frac{1}{n}\cdot \left (1-\frac{1}{n!}\right ) \\ & = \sum_{k=0}^{n}\frac{1}{k!}+\frac{1}{n}-\frac{1}{n}+\frac{1}{n!}\cdot \frac{1}{n} \\ & \geq \sum_{k=0}^{n}\frac{1}{k!}\end{align*}

Therefore we have that \begin{equation*}\left (1+\frac{1}{n}\right )^n\leq \sum_{k=0}^n \frac{1}{k!}\leq \left (1+\frac{1}{n}\right )^{n+1}\end{equation*}

Taking the limit $n\rightarrow \infty$ we get \begin{equation*}e\leq \sum_{k=0}^{\infty} \frac{1}{k!}\leq e \Rightarrow \sum_{k=0}^{\infty} \frac{1}{k!}=e\end{equation*}


Is everything correct? (Wondering)

 

[I like Serena]

It seems correct to me, although that is a lot of steps.

In particular I also see the term $\frac{1}{n!n}$ in the last step that I believe we need for (3)? (Thinking)

 

[mathmari]

It seems correct to me, although that is a lot of steps.

Can we make it a bit shorter? (Wondering)

2. Calculate $s_n$ and $\left (1+\frac{1}{n}\right )^n$ for $n=10$ and $n=100$ with exactly $12$ decimal places and compare with $e$.

According to the calculator we get (with 12 decimal places) $s_{10}=\sum_{k=0}^{10}\frac{1}{k!}=2.718281801146$ and $\left (1+\frac{1}{10}\right )^{10}=2.593742460100$. We have that $e=2.718281828459$. We see that $s_{10}$ is closer to $e$ as $\left (1+\frac{1}{10}\right )^{10}$.

According to the calculator we get (with 12 decimal places) $s_{100}=\sum_{k=0}^{100}\frac{1}{k!}=2.718281828459$ and $\left (1+\frac{1}{100}\right )^{100}=2.704813829422$. We have that $e=2.718281828459$. We see that that the first 12 deimal places of $s_{100}$ are the same as these of $e$.

Is this the comparison? Or do we have to say also something else? (Wondering)

 

[mathmari]

3. Show that the differenz of the $n$-th parial sum of the series $\sum_{k=0}^{\infty}\frac{1}{k!}$ can be estimated to $e$ by $0<e-s_n<\frac{1}{n!n}$.

In particular I also see the term $\frac{1}{n!n}$ in the last step that I believe we need for (3)? (Thinking)

How can we use this term? I got stuck right now.

\begin{equation*}\left (1+\frac{1}{n}\right )^{n+1} \geq \sum_{k=0}^{n}\frac{1}{k!}+\frac{1}{n!}\cdot \frac{1}{n} \end{equation*} The left side is also bigger when $n\rightarrow \infty$. So, we get\begin{align*}&\lim_{n\rightarrow\infty}\left (1+\frac{1}{n}\right )^{n+1} \geq \sum_{k=0}^{n}\frac{1}{k!}+\frac{1}{n!\cdot n}\Rightarrow \lim_{n\rightarrow\infty}\left (1+\frac{1}{n}\right )^{n+1} - \sum_{k=0}^{n}\frac{1}{k!}\geq\frac{1}{n!\cdot n} \\ & \Rightarrow e-s_n\geq\frac{1}{n!\cdot n}\end{align*} In this way we get the wrong inequality.

Or have I done something wrong?

(Wondering)

 

[I like Serena]

Can we make it a bit shorter?

As yet I don't see how. (Worried)

According to the calculator we get (with 12 decimal places) $s_{10}=\sum_{k=0}^{10}\frac{1}{k!}=2.718281801146$ and $\left (1+\frac{1}{10}\right )^{10}=2.593742460100$. We have that $e=2.718281828459$. We see that $s_{10}$ is closer to $e$ as $\left (1+\frac{1}{10}\right )^{10}$.

According to the calculator we get (with 12 decimal places) $s_{100}=\sum_{k=0}^{100}\frac{1}{k!}=2.718281828459$ and $\left (1+\frac{1}{100}\right )^{100}=2.704813829422$. We have that $e=2.718281828459$. We see that that the first 12 deimal places of $s_{100}$ are the same as these of $e$.

Is this the comparison? Or do we have to say also something else?

I believe so yes.
In particular we see that $s_n$ approaches $e$ much faster. (Nerd)

How can we use this term? I got stuck right now.

\begin{equation*}\left (1+\frac{1}{n}\right )^{n+1} \geq \sum_{k=0}^{n}\frac{1}{k!}+\frac{1}{n!}\cdot \frac{1}{n} \end{equation*} The left side is also bigger when $n\rightarrow \infty$. So, we get\begin{align*}&\lim_{n\rightarrow\infty}\left (1+\frac{1}{n}\right )^{n+1} \geq \sum_{k=0}^{n}\frac{1}{k!}+\frac{1}{n!\cdot n}\Rightarrow \lim_{n\rightarrow\infty}\left (1+\frac{1}{n}\right )^{n+1} - \sum_{k=0}^{n}\frac{1}{k!}\geq\frac{1}{n!\cdot n} \\ & \Rightarrow e-s_n\geq\frac{1}{n!\cdot n}\end{align*} In this way we get the wrong inequality.

Or have I done something wrong?

As you said, I think we cannot do it this way after all because the bound is on the wrong side of $e$. (Worried)

Instead we can use for part (3) that $s_n$ is the Taylor series for $e^x$ with $x=1$.
Its Lagrange form of the remainder is $$R_n(1)=\frac{\exp^{(n+1)}(\xi)}{(n+1)!}x^{n+1}\Big|_{x=1}= \dfrac{e^\xi}{(n+1)!}$$
where $0<\xi<x = 1$.
So:
$$
0<\frac{e^0}{(n+1)!} < e - s_n < \frac{e^1}{(n+1)!}
$$
Hmm... it seems we're just a little short of $\frac{1}{n!n}$. (Sweating)