Let's find Euler's number - e

The Euler’s number is another mathematical constant which is an irrational number similar to $latex \pi$. In addition, this is the base of the natural logarithm. Few of the interesting things related to this number are as follows;

• $latex \displaystyle e=\lim_{n\rightarrow\infty} \bigg(1+\frac{1}{n}\bigg)^n$.
• $latex \displaystyle e = \sum_{n=0}^\infty {\frac{1}{n!}}$.
• When $latex f(x)=e^x$, the first derivation of the function (gradient of $latex f$ at point $latex x$) is $latex f^1(x)=e^x$. Moreover, for any integer $latex n$ (positive for differentiation and negative for integration neglecting the arbitrary constant), the n-th derivation is $latex f^n(x)=e^x$.
• As $latex x\rightarrow\infty$, $latex e^x\rightarrow\infty$ and as $latex x\rightarrow-\infty$, $latex e^x\rightarrow0$. And $latex e^0=1$ which is an obvious fact. The function $latex e^{-x}$ is the reflection of $latex e^x$ around the y-axis.

Today we are going to use a simple technique to estimate the value of ‘$latex e$’ (You probably can search the value from the internet easily). We are going to use the plots in the figure below.

We have $latex y=e^{-x}$ curve and it is bounded by two step functions. In the step functions, the width of each column (step size) is $latex \frac{1}{n}$, i.e. the unit length of x-axis is divided in to '$latex n$' segments. 

Let's look at the two overlapped columns starting from point '$latex x$'.

• Height of the tall column is $latex e^{-x}$.
• Height of the short column is $latex e^{-(x+\frac{1}{n})}$.
• Width of both columns are $latex \frac{1}{n}$.

Therefore, the surface area of the tall column is $latex \frac{1}{n}\times{e^{-x}}$ and for the short column it is $latex \frac{1}{n}\times{e^{-(x+1/n)}}=\frac{e^{-1/n}}{n}\times{e^{-x}}$.

With the knowledge of calculus, we can say that the area under the curve '$latex y$' within the width of those columns is $latex \int_x^{(x+1/n)}e^{-x}dx$. 

We can observe that the area under the curve is larger than the area of short column, but smaller than the area of the tall column. Thus,

$latex \frac{e^{-1/n}}{n}{e^{-x}}<\int_x^{(x+1/n)}e^{-x}dx<\frac{1}{n}{e^{-x}}$

Now, let us add together all the areas starting from $latex x=1$ to the infinity by the step size of $latex \Delta{x}={1}/{n}$,

$latex \displaystyle \sum_{x=1,\Delta{x}=\frac{1}{n}}^{\infty}\frac{e^{-1/n}}{n}{e^{-x}}<\displaystyle\sum_{x=1,\Delta{x}=\frac{1}{n}}^{\infty}\int_x^{(x+1/n)}e^{-x}dx <\displaystyle\sum_{x=1,\Delta{x}=\frac{1}{n}}^{\infty}\frac{1}{n}{e^{-x}}$

$latex \frac{e^{-1/n}}{n}\displaystyle\sum_{x=1,\Delta{x}=\frac{1}{n}}^{\infty}{e^{-x}}<\int_1^{\infty}e^{-x}dx<\frac{1}{n}\displaystyle\sum_{x=1,\Delta{x}=\frac{1}{n}}^{\infty}{e^{-x}}$

If you look at the infinite summations, you may notice that both of them are geometric progressions. Right series has $latex \frac{e^{-1}}{n}$ as the first term and for the left series it is $latex \frac{e^{-1}e^{-1/n}}{n}$ and, both of them have $latex e^{-1/n}$ as the common ratio (it is less than 1 which we take by the fact of that '$latex e$' is larger than 1).

Thus, by the knowledge of infinite sum of geometric series and integration, we can re-write above inequalities as below.

$latex \frac{(e^{-1}e^{-1/n})/n}{1-e^{-1/n}}<e^{-1}<\frac{(e^{-1})/n}{1-e^{-1/n}}$

$latex \frac{e^{-1/n}}{n}<1-e^{-1/n}<\frac{1}{n}$

$latex \frac{1}{n}<e^{1/n}-1<\frac{e^{1/n}}{n}$

By considering left-middle terms and middle-right terms as separate inequalities, we can simplify above furthermore and later combine it as follows;

$latex \frac{n+1}{n}<e^{1/n}<\frac{n}{n-1}$

$latex (\frac{n+1}{n})^n<e<(\frac{n}{n-1})^n$

There it is. By using any positive integer for '$latex n$', we can estimate the value of '$latex e$'.

Examples: 

1. Let $latex n=8$. Then,

$latex (\frac{8+1}{8})^4<e<(\frac{8}{8-1})^8$

$latex 2.56578<e<2.91028$

However, this approach converges very slowly. Let's say that we need to find the exact answer for $latex k$ decimal points. That implies $latex (\frac{n}{n-1})^n-(\frac{n+1}{n})^n<10^{-k}$ need to be satisfied. Now we are going to simplify this.

$latex (\frac{n}{n-1})^n-(\frac{n+1}{n})^n<10^{-k}$

$latex (\frac{n^2}{n^2-1})^n-1<(\frac{n}{n+1})^n10^{-k}$

We know that as '$latex n$' increases the lower bound increases. Therefore, the reciprocal decreases. From above example we can guess that the lower bound never reaches to 3 which means the reciprocal never drops below 1/3. So above inequality is simplified with that assumption ( $latex \forall{n\in\mathbb{Z}^+},(\frac{n}{n+1})^n<\frac{1}{3}$ ).

Therefore if we can find an '$latex n$' which satisfy $latex (\frac{n^2}{n^2-1})^n-1<\frac{1}{3}10^{-k}$ for given '$latex k$', using that '$latex n$' we can estimate the value of '$latex e$' for '$latex k$' decimal points correctly.

Examples: 

1. Let $latex n=100$. Then $latex 3[(\frac{n^2}{n^2-1})^n-1]=0.03015<10^{-1}$. Therefore, $latex (\frac{100+1}{100})^{100}=2.7048$ gives the estimation for one decimal point, which is $latex e\simeq2.7$.
2. Let $latex n=1000$. Then $latex 3[(\frac{n^2}{n^2-1})^n-1]=0.0030015<10^{-2}$. Therefore, $latex (\frac{100+1}{100})^{100}=2.7169$ gives the estimation for two decimal point, which is $latex e\simeq2.71$.
3. Let $latex n=10000$. Then $latex 3[(\frac{n^2}{n^2-1})^n-1]=0.000300015<10^{-3}$. Therefore, $latex (\frac{1000+1}{1000})^{1000}=2.71814$ gives the estimation for three decimal point, which is $latex e\simeq2.718$.

Hope you got the idea. And have fun with calculating to large number of decimal points.

From wikipedia I got it for 50 decimal points, which is

e = 2.71828182845904523536028747135266249775724709369995.

The infinite sum of 1/xⁿ

”Convergence of series” is a quite interesting topic during high school mathematics. A popular such series is $latex \sum_{x\in\mathbb{Z}^+}(\frac{1}{x^n})$ where $latex n$ is a positive integer. When $latex n>1$, there were lots of examples to determine the convergence and finding the sum up to general $latex N$ terms.

However, it is not true for $latex n=1$ and I always wondered how it happens. Therefore, I thought to share the proof with a simple graphical approach as follows.

Let’s look at the figure.

We have two curves: $latex y_1=\frac{1}{x^n}$ and $latex y_2=\frac{1}{(x+1)^n}$. In between them, I drew a bar-graph and you may notice that the height of each column is equivalent to $latex y=\frac{1}{{\lfloor x\rfloor}^n}$ where $latex \lfloor x\rfloor$ is the closest lower integer of $latex x$.
( Eg: $latex \lfloor 2.5 \rfloor = \lfloor 2 \rfloor = 2$ )

Now, if we try to calculate the area of $latex k$-th column, it will be $latex \frac{1}{k^n}\times 1=\frac{1}{k^n}\$. Therefore, what will be the sum of area of all columns? Yes, it is our nice buddy series, $latex \sum_{x=1}^{\infty}(\frac{1}{x^n})$. Pretty interesting, isn’t it?

Now look at the area under each curve for positive $latex x$ values.

We can see that each column is popping out a little bit from the curve $latex y_2$. Therefore, we can say that the area under curve $latex y_2$ is less than the sum area of columns.

And it is the opposite for curve $latex y_1$. Every column is underneath the curve and thus its area is larger than the sum area of columns. We know that $latex y_1$ tends to infinity when it is close to 0. Thus, we modify our argument like this. For $latex x>1$, we can say that the area under $latex y_1$ is larger than the sum area of columns excluding the first column. The area of the 1^st column is 1. Then, we can say the sum area of columns is less than the area under the curve $latex y_1$ for $latex x>1$ plus 1.

(Area under $latex y_2$ for $latex x>0$ ) < $latex \sum_{x=1}^{\infty}(\frac{1}{x^n})$ < (1 + Area under $latex y_1$ for $latex x>1$ )

The guys who play with calculus know that the area under a given curve $latex f(x)$ between the interval $latex [a,b]$ is $latex \int_a^b|f(x)|dx$. Therefore, we can make above relation in a mathematical form as following; 

$latex \int_0^\infty\frac{1}{(x+1)^n}dx < \sum_{x>0}(\frac{1}{x^n}) <1 + \int_1^\infty\frac{1}{x^n}dx$,

where $latex x \in \mathbb{Z}^+$.

Let’s solve above relations.

• For $latex n=1$,

$latex \int_0^\infty\frac{1}{(x+1)}dx < \sum_{x>0}(\frac{1}{x}) < 1 +\int_1^\infty\frac{1}{x}dx$

$latex \ln(x+1)|_0^\infty < \sum_{x>0}(\frac{1}{x}) <1 + \ln(x)|_1^\infty$

$latex \infty < \sum_{x>0}(\frac{1}{x}) < \infty$

Yeah, $latex \sum_{x>0}(\frac{1}{x})$ is not going to converge.

• For $latex n > 1$,

$latex \int_0^\infty\frac{1}{(x+1)^n}dx < \sum_{x>0}(\frac{1}{x^n}) <1 + \int_1^\infty\frac{1}{x^n}dx$

$latex \frac{1}{(1-n)(x+1)^{n-1}}|_0^\infty < \sum_{x>0}(\frac{1}{x^n}) < 1 + \frac{1}{(1-n)x^{n-1}}|_1^\infty$
 $latex \frac{1}{(n-1)} < \sum_{x > 0}(\frac{1}{x^n}) < \frac{n}{(n-1)} $

As $latex n\rightarrow\infty$,

$latex 0 < \sum_{x > 0}(\frac{1}{x^n}) < 1 $

Series converges.

Furthermore, for $latex n < 1$ and $latex \forall x \in \mathbb{Z}^+$, 

$latex \frac{1}{x} <\frac{1}{x^n} $

$latex \sum_{x>0}(\frac{1}{x} )<\sum_{x>0}(\frac{1}{x^n})$

          $latex \infty < \sum_{x>0}(\frac{1}{x^n})$

Therefore, we can conclude that,

• For $latex n > 1$ ; $latex \sum_{x=1}^{\infty}(\frac{1}{x^n})$ converges.
• For $latex n \le 1$ ; $latex \sum_{x=1}^{\infty}(\frac{1}{x^n})$ diverges.