By the intuition from our birth, we love summing things, categorising the similar things and so on. Consider a sum $A=1+2+3+4+\dots$ what we have here is infinite sum, as “$\dots$” imply “this sum goes to infinity with the order you see”. So in a some sense, it behaves like sequences, because we write the things in order like $1$, $2$, $3$, $4$, $\dots$ and then we sum them. Mathematicians tought that so, and as a result they came up with a new topic called “series”, especially “infinite series”. Spoiler alert, an infinite series can converge, if not we say series is divergent.

Infinity and Beyond!

$\textbf{\small Definition 1.18.:} $ (Infinite Series) Let $(b_n)$ be a sequence. An infinite series is a formal expression of the form

$$ \begin{aligned} \sum_{n=1}^{\infty} b_n = b_1+b_2+\dots \end{aligned} $$
$$ \begin{aligned} \sum_{n=1}^{\infty} b_n = b_1+b_2+\dots \end{aligned} $$

So as we said, it can diverge or converge.

$\textbf{\small Definition 1.19.:} $ (Convergence of Series) Let $(s_m)$ be a sequence of partial sums of a series $\sum_{n=1}^{\infty} b_n$, which written as

$$ \begin{aligned} s_m = b_1+b_2+\dots+b_m \end{aligned} $$
$$ \begin{aligned} s_m = b_1+b_2+\dots+b_m \end{aligned} $$

and say that the series $\sum_{n=1}^{\infty} b_n$ converges $B$ if the sequence $(s_m)$ converges to $B$. In this case we write $\sum_{n=1}^{\infty} b_n = B$.

It is intuitive to think, if an infinite series converges, then finite sum of the series also converges.

$\textbf{\small Example:} $ Consider the following very famous example. Check that if

$$ \begin{aligned} \sum_{n=1}^{\infty} \dfrac{1}{n^2} \end{aligned} $$
$$ \begin{aligned} \sum_{n=1}^{\infty} \dfrac{1}{n^2} \end{aligned} $$

converges

$\textbf{\small Solution:} $ Take the partial sum $s(m)$

$$ \begin{aligned} s_m = 1 + \dfrac{1}{4}+\dfrac{1}{9}+\dots+\dfrac{1}{m^2} \end{aligned} $$
$$ \begin{aligned} s_m = 1 + \dfrac{1}{4}+\dfrac{1}{9}+\dots+\dfrac{1}{m^2} \end{aligned} $$

as definition of convergence of series implies if $(s_m)$ converges to somewhere, then we say the series also converges. We will use monotone convergence theorem. So, first we will show that the partial sum $(s_m)$ is monotone and then bounded.

Consider,

$$ \begin{aligned} s_{m+1} - s_m &= \dfrac{1}{(m+1)^2} - \dfrac{1}{m^2} \\ &= \dfrac{m^2 - m^2 - 2m - 1}{m^2(m+1)^2} \\ &= -\left(\dfrac{2m+1}{(m(m+1))^2}\right) < 0 \end{aligned} $$
$$ \begin{aligned} s_{m+1} - s_m &= \dfrac{1}{(m+1)^2} - \dfrac{1}{m^2} \\ &= \dfrac{m^2 - m^2 - 2m - 1}{m^2(m+1)^2} \\ &= -\left(\dfrac{2m+1}{(m(m+1))^2}\right) < 0 \end{aligned} $$

Thus,

$$ \begin{aligned} s_{m+1} < s_m \end{aligned} $$
$$ \begin{aligned} s_{m+1} < s_m \end{aligned} $$

$s_m$ is monotone decreasing sequence. Now let us find a bound for partial sum

$$ \begin{aligned} s_m &= 1 + \dfrac{1}{4} + \dfrac{1}{9} + \dots + \dfrac{1}{m^2}\\ &<1 +\dfrac{1}{2\cdot 2} + \dfrac{1}{3\cdot 3} + \dots +\dfrac{1}{m\cdot m} \\ &<1 +\dfrac{1}{2\cdot 1} + \dfrac{1}{3\cdot 2} + \dots +\dfrac{1}{m\cdot (m-1)} \\ &=1 +\left(1-\dfrac{1}{2}\right) + \left(\dfrac{1}{2}-\dfrac{1}{3}\right) + \dots+ \left(\dfrac{1}{(m-1)} - \dfrac{1}{m}\right)\\ &= 1+1-\dfrac{1}{m} < 2 \end{aligned} $$
$$ \begin{aligned} s_m &= 1 + \dfrac{1}{4} + \dfrac{1}{9} + \dots + \dfrac{1}{m^2}\\ &<1 +\dfrac{1}{2\cdot 2} + \dfrac{1}{3\cdot 3} + \dots +\dfrac{1}{m\cdot m} \\ &<1 +\dfrac{1}{2\cdot 1} + \dfrac{1}{3\cdot 2} + \dots +\dfrac{1}{m\cdot (m-1)} \\ &=1 +\left(1-\dfrac{1}{2}\right) + \left(\dfrac{1}{2}-\dfrac{1}{3}\right) + \dots+ \left(\dfrac{1}{(m-1)} - \dfrac{1}{m}\right)\\ &= 1+1-\dfrac{1}{m} < 2 \end{aligned} $$

And we see that $2$ is an upper bound for $s_m$. By the monotone convergence theorem, $\sum_{n=1}^{\infty} \frac{1}{n^2}$ converges to a limit less than $2$. Later, we will show the exact value of the limit.

$\textbf{\small Theorem 1.12.:} $ (Cauchy Convergence Criteria) The series $\sum_{n=1}^{\infty} a_k$ is convergent if and only if $\varepsilon > 0$, $\exists N(\varepsilon)$, $\forall n > N$

$$ \begin{aligned} \vert a_{m+1} + a_{m+2} + \dots + a_{n} \vert < \varepsilon \end{aligned} $$
$$ \begin{aligned} \vert a_{m+1} + a_{m+2} + \dots + a_{n} \vert < \varepsilon \end{aligned} $$

$\textbf{\small Proof:}$ Let us begin with ($\Rightarrow$) side. Suppose that $\sum_{n=1}^{\infty} a_k$ converges. This means the partial sum $s_m = a_1 + a_2 + \dots + a_m$ converges and therefore, $s_m$ is Cauchy sequence.

Let $\varepsilon > 0$ be given, then there exists $N(\varepsilon)$, $\forall n,m > N$. Take,

$$ \begin{aligned} \vert s_n-s_m \vert &= \vert a_{m+1} + a_{m+2} + \dots + a_n \vert \\ \vert s_n-s_m \vert &< \varepsilon \end{aligned} $$
$$ \begin{aligned} \vert s_n-s_m \vert &= \vert a_{m+1} + a_{m+2} + \dots + a_n \vert \\ \vert s_n-s_m \vert &< \varepsilon \end{aligned} $$

Now let us prove the other ($\Leftarrow$) side. Let $s_m = a_{1} + a_{2} + \dots + a_{m}$ be a convergent sequence. Then $s_n$ is a Cauchy sequence for all $n,m$. WLOG, $n > m$. Consider,

$$ \begin{aligned} \vert s_n-s_m \vert &= \vert a_{m+1} + a_{m+2} + \dots + a_n \vert \end{aligned} $$
$$ \begin{aligned} \vert s_n-s_m \vert &= \vert a_{m+1} + a_{m+2} + \dots + a_n \vert \end{aligned} $$

and also since convergent sequence $s_m$ is a partial sum of $\sum_{n=1}^{\infty} a_k$. Means that, the sum $\sum_{n=1}^{\infty} a_k$ also converge. Now, let us investigate harmonic series

$\textbf{\small Example:} $ Show that

$$ \begin{aligned} \sum_{n=1}^{\infty} \dfrac{1}{n} \end{aligned} $$
$$ \begin{aligned} \sum_{n=1}^{\infty} \dfrac{1}{n} \end{aligned} $$
is divergent.

$\textbf{\small Solution:} $ Consider the following term

$$ \begin{aligned} \vert s_{2n} - s_n \vert &= \left\vert 1 + \dfrac{1}{2} + \dots \dfrac{1}{n} + \dfrac{1}{n+1} + \dots + \dfrac{1}{2n} - 1 - \dfrac{1}{2} - \dots - \dfrac{1}{n}\right\vert \\ &=\left\vert \dfrac{1}{n+1} +\dfrac{1}{n+2} + \dots + \dfrac{1}{2n} \right\vert \\ &>\dfrac{1}{2n} +\dfrac{1}{2n} + \dots \dfrac{1}{2n} \\ &> n \cdot \dfrac{1}{2n} = \dfrac{1}{2} \end{aligned} $$
$$ \begin{aligned} \vert s_{2n} - s_n \vert &= \left\vert 1 + \dfrac{1}{2} + \dots \dfrac{1}{n} + \dfrac{1}{n+1} + \dots + \dfrac{1}{2n} - 1 - \dfrac{1}{2} - \dots - \dfrac{1}{n}\right\vert \\ &=\left\vert \dfrac{1}{n+1} +\dfrac{1}{n+2} + \dots + \dfrac{1}{2n} \right\vert \\ &>\dfrac{1}{2n} +\dfrac{1}{2n} + \dots \dfrac{1}{2n} \\ &> n \cdot \dfrac{1}{2n} = \dfrac{1}{2} \end{aligned} $$

Hence, $s_m$ is not Cauchy sequence and by theorem 1.12 we conclude that the sum $\sum_{n=1}^{\infty} \dfrac{1}{n}$ is divergent.

Remember that, Cauchy sequence says “there is a limit that this sequence goes, but I do not know what it is.”, similarly the theorem 1.13 reads that “if the sum of many terms of a series is smaller than arbitrary small number, then you can say the series converges.”.

$\textbf{\small Theorem 1.13.:} $ If $\sum_{n=1}^{\infty} a_n$ converges, then $\lim_{n\rightarrow \infty} a_n = 0$

$\textbf{\small Proof:}$ Take partial sum $s_n = a_1 + a_2 + \dots + a_n$. Also take, $s_{n-1} = a_1 + a_2 + \dots + a_{n-1}$. If $\sum a_n$ converges, say $s$, we can say that partial sum $s_m$ also converges by definition 1.19.

$$ \begin{align} \lim_{n \rightarrow \infty} &(s_n - s_{n-1}) \notag \\ = &(s-s) \notag \\ = &0 \end{align} $$
$$ \begin{align} \lim_{n \rightarrow \infty} &(s_n - s_{n-1}) = (s-s) = 0 \end{align} $$

and also

$$ \begin{align} &s_n-s_{n-1} \notag \\ =&a_1 + a_2+ \dots + a_{n-1} + a_n \notag \\ -&a_1 - a_2 - \dots - a_{n-1} \notag \\ =&a_n \end{align} $$
$$ \begin{align} &s_n-s_{n-1} = a_1 + a_2+ \dots + a_{n-1} + a_n - a_1 - a_2 - \dots - a_{n-1} = a_n \end{align} $$

Using $(1)$ and $(2)$, we complete the proof

$$ \begin{align} \lim_{n \rightarrow \infty} a_n = 0\notag \end{align} $$
$$ \begin{align} \lim_{n \rightarrow \infty} a_n = 0\notag \end{align} $$

Observe that $a_n$ in above theorem is the very sequence used in the series, not the partial sum of the series.

Therefore, converse of the theorem 1.13 is not true. We cannot guarantee if $\lim_{n \rightarrow \infty} a_n = 0$, then $\sum a_k$ converges. Most famous example for that is harmonic series. Because, $\lim_{n \rightarrow \infty} \dfrac{1}{n} = 0$, but $\sum_{n = 1}^{\infty} \dfrac{1}{n}$ diverges.

Last but not least, contrapositive of it is very handy when it comes to deciding whether a series diverges. Note that if $\lim_{n \rightarrow \infty} a_n \neq 0$, then $\sum_{n=1}^{\infty} a_n$ diverges. For instance, $\sum_{n=1}^{\infty} \cos\left(\dfrac{1}{n}\right)$ diverges because $\lim_{n \rightarrow \infty} \cos\left(\dfrac{1}{n}\right) \neq 0$.

As similar to the sequences, series also has tail. Rigorously,

$$ \begin{aligned} \sum_{n=M}^{\infty} a_n \end{aligned} $$
$$ \begin{aligned} \sum_{n=M}^{\infty} a_n \end{aligned} $$

$\textbf{\small Lemma 1.3.:}$ The series $\sum_{n=1}^{\infty} a_n$ converges if and only if $\sum_{n=M}^{\infty} a_n$ converges for any natural number $M$.

Now, we will define a very fundamental, so-called geometric series that is useful from geometry to functional analysis.

Geometric Series

$\textbf{\small Definition 1.20.:} $ (Geometric Series) An infinite series

$$ \begin{aligned} \sum_{n=1}^{\infty} ar^{n-1} = a + ar + ar^2 + \dots \end{aligned} $$
$$ \begin{aligned} \sum_{n=1}^{\infty} ar^{n-1} = a + ar + ar^2 + \dots \end{aligned} $$

is called geometric series. $r$ is called common ratio. Let us investigate the properties.

If $\vert r \vert < 1$, then $\lim_{n \rightarrow \infty} r^n = 0$. Take partial sum $S_n = a_1 + a_2 + a_3 + \dots + a_n = a + ar + ar^2 + \dots + ar^{n-1}$. Also observe that, $r\cdot S_n = ar + ar^2 + ar^3 + \dots + ar^{n}$. Thus,

$$ \begin{align} S_n - rS_n = &a + ar + ar^2 + \dots + ar^{n-1} \notag\\ - &ar - ar^2 - ar^3 - \dots - ar^{n} \notag\\ = &a - ar^{n} \notag\\ S_n(1-r) = &a - ar^{n} \notag\\ S_n = &a \cdot \left(\dfrac{1-r^{n}}{1-r}\right) \end{align} $$
$$ \begin{align} S_n - rS_n = &a + ar + ar^2 + \dots + ar^{n-1} \notag\\ - &ar - ar^2 - ar^3 - \dots - ar^{n} \notag\\ = &a - ar^{n} \notag\\ S_n(1-r) = &a - ar^{n} \notag\\ S_n = &a \cdot \left(\dfrac{1-r^{n}}{1-r}\right) \end{align} $$

By definition, we know if partial sum $S_n$ converges to a limit, then the series $\sum ar^{n-1}$ also converges. Hence, it is enough to evaluate the limit of the sequence $S_n$.

$$ \begin{aligned} &\lim_{n\rightarrow \infty} S_n \\ = &\lim_{n \rightarrow \infty} a \cdot \left(\dfrac{1-r^{n+1}}{1-r}\right)\\ = &\dfrac{a}{1-r} \end{aligned} $$
$$ \begin{aligned} &\lim_{n\rightarrow \infty} S_n \\ = &\lim_{n \rightarrow \infty} a \cdot \left(\dfrac{1-r^{n+1}}{1-r}\right)\\ = &\dfrac{a}{1-r} \end{aligned} $$

So, we conclude that the series $\sum_{n=1}^{\infty} ar^{n-1}$ converges. Geometric series diverge for the cases where $\vert r \vert \geq 1$ and $\vert r \vert = 1$.

Let us do two examples:

$\textbf{\small Example:} $ Determine whether the following series is convergent or divergent.

$$ \begin{aligned} \sum_{n=1}^{\infty} 3^{2k}5^{1-k} \end{aligned} $$
$$ \begin{aligned} \sum_{n=1}^{\infty} 3^{2k}5^{k-1} \end{aligned} $$

$\textbf{\small Solution:} $ Rearrange the series as

$$ \begin{aligned} \sum_{n=1}^{\infty} 9\cdot \left(\dfrac{9}{5}\right)^{k-1} \end{aligned} $$
$$ \begin{aligned} \sum_{n=1}^{\infty} 9\cdot \left(\dfrac{9}{5}\right)^{k-1} \end{aligned} $$
Since $\dfrac{9}{5} > 1$ we say that the series diverges. Now let us examine the next example

$\textbf{\small Example:} $ Determine whether the following series is convergent or divergent.

$$ \begin{aligned} \sum_{n=1}^{\infty} \dfrac{5}{4^n } \end{aligned} $$
$$ \begin{aligned} \sum_{n=1}^{\infty} \dfrac{5}{4^n } \end{aligned} $$

$\textbf{\small Solution:} $ It is trivial to see that $\dfrac{1}{4} < 1$ and hence the series converges. Furthermore, being $a=5$ and $r=\dfrac{1}{4}$ would give us:

$$ \begin{aligned} \sum_{n=1}^{\infty} \dfrac{5}{4^n } = \dfrac{5}{1-\dfrac{1}{4}} = \dfrac{20}{3} \end{aligned} $$
$$ \begin{aligned} \sum_{n=1}^{\infty} \dfrac{5}{4^n } = \dfrac{5}{1-\dfrac{1}{4}} = \dfrac{20}{3} \end{aligned} $$

For now, let us call it a post. In the next section we will talk about telescoping series, algebraic properties of series and other convergence tests. Take care!