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.