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Infinite series

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Definition

is an expression of the form:

$$ \sum_{n=1}^{\infty} a_n = a_1 + a_2 + \dots + a_i + \dots $$

The $N^{th}$partial sum $S_n$is the sum of the first N terms:

$$ S_n = a_1 + a_2 + \dots + a_N $$

We say the infinite series $\sum_{n=1}^{\infty} a_n$is convergent with sum $S$provided:

$$ S = \lim_{N \to \infty} S_N $$

if $\lim_{N \to \infty} S_N$does not exist, we say that the series diverges.

Examples

I'll provide an example of an infinite series and explain its convergence or divergence using latex notation.

Let's consider the geometric series with first term $\frac{1}{2}$:

$$ \sum_{n=1}^{\infty} \frac{1}{2^n} = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots $$

1) First, let's find the Nth partial sum $S_N$:

$$ S_N = \frac{1}{2} + \frac{1}{4} + ... + \frac{1}{2^{N}} $$

2) This is a geometric series with first term $a=1$ and ratio $r=\frac{1}{2}$.

  • $a$ is the first term
  • $r$ is the common ratio ($|r| < 1$for convergence)
  • $N$ is the number of terms in the partial sum

First term: geometric series is simply value of the first number in the sequence

Common ratio: constant factor by which each term in the series is multiplied to get the next term by dividing any term in the series by the term currently before it:

$$ r = \frac{\text{next term}}{\text{current term}} $$

The formula for the partial sum of a geometric series: $S_n = a\frac{1-r^N}{1-r}$

Then apply below:

$$ S_N = \frac{a(1-r^N)}{1-r} = \frac{1}{2}\frac{1-(\frac{1}{2})^N}{1-\frac{1}{2}} = 1-(\frac{1}{2})^N $$

3) To find if the series converges, we take the limit as N approaches infinity:

When $N \to \infty$, the term $\left( \frac{1}{2} \right)^N$(which is $\frac{1}{2^N}$) approaches 0 because $2^N$grows infinitely large.

$$ \lim_{N \to \infty} S_N = \lim_{N \to \infty} 2(1-(\frac{1}{2})^N) = 1-0 = 1 $$

Since this limit exists and equals 2, we can say that:

$$ \sum_{n=1}^{\infty} \frac{1}{2^n} = 1 $$

Therefore, this infinite series converges to 1.

p-series

The p-series $\sum_{n=1}^{\infty} \frac{1}{n^p}$is convergent if $p > 1$and divergent if $p \le$ 1

There are two kinds of series:

  • Convergence: Series converges if $p > 1$, which means that as you increases the series for $p > 1$, therefore, the sum approaches a finite value.
    – The reason is when $p > 1$, the terms $\frac{1}{n^P}$ decreased quickly enough as $n \to \infty$, so the total sum remains finite
  • Divergence: The series diverges if $p \le 1$. This means that as keeping decreasing p from 1, the sum grows infinitely large or does not settle to a finite value.
    – The reason when $p \le 1$, the terms $\frac{1}{n^P}$decrease too slowly (or not at all when $p = 0$), so the sum rises without bound.

Example:

  • With p = 2: $\sum_{n=1}^{\infty} \frac{1}{n^2}$, and $p = 2 > 1$means that this converges to a finite value $\frac{\pi^2}{6}$
  • With p = 1: $\sum_{n=1}^{\infty} \frac{1}{n}$, and $p = 1 \le 1$means that this is the harmonic series, which diverges.

Geometric series

In the form

$$ \sum_{n=0}^{\infty} x^n = 1 + x + x^2 + x^3 + \dots $$

  • Converges if $|x| < 1$, the reason is when $|x| < 1$, each term $x^n$gets smaller and smaller as $n \to \infty$, so the sum approaches a finite value:

$$ \sum_{n = 0}^{\infty} x^n = \frac{1}{1-x}, \space for |x| < 1 $$

  • Divergence: The series diverges if $x \ge 1$, the reason is when $|x| \ge 1$, the terms $x^n$ either do not decrease or grow larger, causing the sum to either oscillate indefinitely (if $x = -1$) or grow infinity large

Examples

Convergences:

When $x = \frac{1}{2}$

$\sum_{n=0}^{\infty} \left( \frac{1}{2} \right)^n = 1 +\frac{1}{2} +\frac{1}{4} + \frac{1}{8} + \dots$which make this converges to $\frac{1}{1 - \frac{1}{2}} = 2$, also we can do same approach on the difference of the current and next term below:

$$ r = \frac{\frac{1}{2}}{1} = 1/2, \frac{\frac{1}{8}}{\frac{1}{2}} = \frac{1}{4} $$

Divergences

When $x = 2$

$\sum_{n=0}^{\infty} 2^n = 1 + 2 + 4 + 8 + \dots$This makes the sum of the series grows infinitely large.

When $x = -1$

$\sum_{n=0}^{\infty} (-1)^n = 1 - 1 + 1 - 1 + \dots$This makes the sum of the series oscillates between 0 and 1, never setting to a finite value. This will leads to the divergence of the Geometric series with $x = -1$

Practice

$\sum_{n=1}^{\infty} (0.3)^n$, $0.3 < 1$means it convergences

$\sum_{n=1}^{\infty} \frac{1}{n^{1.1}}$, $1.1 > 1$ means it is convergences

$\sum_{n=1}^{\infty} 2^n$, $2 \ge 1$which means it is divergence

$\sum_{n=1}^{\infty} \frac{1}{n^{0.3}}$, p = 0.3 and $p \le 1$, which means it is divergence