A series with finite terms results a definite sum. However, it is difficult to predict the algebraic sum of an infinite series. It is not necessary that the algebraic sum of all the infinite series is infinite. Some series might lead to a finite sum. That is, it tends to ‘converge’ to a finite number. This is called convergence of the series. **For example**, the reciprocals of factorials of all positive integers from 0 converge as ‘e’, the exponential constant.

However, there are series that do not converge and they are called as ‘divergent’ series. For example, sum of all natural numbers leads to infinity and hence it is a divergent series.

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If we have a series $\sum x_{n}$ which is an infinite series, then we can define a sequence for that as follows:

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$S_n$ = $x_1$ + $x_2$_{ }+ .....+ $x_n$, and so on.

Then, the sequence <$S_n$> is known as the sequence of partial sums of the given series $\sum x_{n}$. The series $\sum x_{n}$ is called convergent, if sequence <$S_n$> is convergent and $\lim_{n\rightarrow \infty }S_{n}$ is known as sum of the given series $\sum x_{n}$. And, if the sequence <$S_n$>is divergent, then the series $\sum x_{n}$ is said to be divergent.

The following are some of the important series which are Convergent.**Convergent Sequence: **A sequence <$a_n$> is said to converge to any number L given any $\epsilon$ > 0, there exists a positive integer $m$ such that $\left | a_{n} - L \right |< \epsilon \forall n \geq m$. The number L is called the limit of the sequence <a_{n}> and is written as $\lim_{n\rightarrow \infty a_{n}}$ = $L$ or $lim a_{n}= L$ or $a_{n}\rightarrow L$.

**Divergent Sequence:** A sequence <$a_n$> is said to diverge or non-convergent to +$\infty$, denoted as $\lim_{n\rightarrow \infty a_{n}}$ = + $\infty$, if to each positive number G, however large, there exists a positive integer m such that, a_{n} > G $\forall n \geq m$.

The given sequence <$S_n$> is close to any number as $n \rightarrow \infty$, then we can say that the given series is converge to any number or limit L and denoted by $\sum_{n = 1}^{\infty}a_{n}$ = $\lim_{n\rightarrow \infty }$$S_{n} = L$

We know that the total sum of any series is the limit of the sequence of respective series. So, we say that if $\sum a_{n}$ is the series and the respective sequence is given by <$S_n$>. Then, the given series $\sum a_{n}$ is convergent iff the sequence <$S_n$> is convergent and we can write it as $\lim_{n\rightarrow \infty }$$S_{n}$ = $\sum_{n = 1}^{\infty}a_{n}$

Let us discuss here two most common convergence theorems.

**Bounded Convergence Theorem**

According to this theorem, if $f_{n}$ be given a a sequence of functions that are measurable and defined over a given measurable set M. If $f_{n} \rightarrow f$ is a pointwise function on set M and $f_{n}$ is bounded uniformly bounded, in other words if $|f_{n}(x)|\ \leq \ M$ for all x and n $\in$ N then $lim \int f_{n} = \int f$**Lebesgue's Dominated Convergence Theorem**

Understand the concept of convergence and divergence with the help of examples.

**Example 1 :** Determine whether the following series converges or diverges : $\sum_{n = 1}^{\infty} n$

**Solution :** We may write this series in terms of sequence of partial sums as :

**Example 2 :** Find if the series $\sum_{n=0}^{\infty}(-1)^{n}$ converges or diverges.

**Solution :** In this case, let us write first few partial sums and check for the nature of series.

**Example 3 :** Check for the following series to be convergent of divergent :

$\sum_{p \geq 0} \frac{4^{2p- 1}}{3^{3p+ 1}}$

**Problem 1 :** Show that the series $\sum_{n \geq 1} \ln \frac{n+1}{n}$ is divergent.

**Problem 2 :** Check for the convergence of following series. If it is convergent find its sum : $\sum_{k \geq 0} \frac{5^{k}+ 4^{k}}{6^{k}}$

**Problem 3 :** Prove that the series $\sum_{n=1}^{\infty} \frac{1}{3^{n-1}}$ is convergent. Also, find its sum.

**Problem 4 :** Test for the convergence of the following series $\sum_{k \geq 1} \frac{(3k)! + 4^{n+1}}{(3k + 1)!}$

$S_1$ = $x_1$

$S_2$ = $x_2$

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$S_n$ = $x_1$ + $x_2$

Then, the sequence <$S_n$> is known as the sequence of partial sums of the given series $\sum x_{n}$. The series $\sum x_{n}$ is called convergent, if sequence <$S_n$> is convergent and $\lim_{n\rightarrow \infty }S_{n}$ is known as sum of the given series $\sum x_{n}$. And, if the sequence <$S_n$>is divergent, then the series $\sum x_{n}$ is said to be divergent.

A simple way of testing a series for convergence or divergence is the ratio test. Compare the unknown series with a known convergent series. Let p_{n} be the n^{th} term of a convergent series. The series with the n^{th} term as $q_n$ will be convergent if 0 < $q_n$_{ }< $p_n$

**For example**, 1 + (1/1) + (1/2) + (1/6) + (1/12) + …..is convergent. Therefore, the series

(1) + (1/2) + (1/4) + (1/12) + (1/24) + ……. will also be convergent.

A geometric infinite series with a common ratio ‘r’ is convergent if r < 1 and divergent if r > 1.

The following are some of the important series which are Convergent.

- The series $p_n$ = 1/(n)(-1)$^{n-1}$ converges to ln 2
- The series $p_n$ = 1/(2n -1)(-1)$^{n-1}$ converges to ($\frac{\pi}{4}$) (Leibnitz’s formula)
- The series $p_n$ = 1/(2$^{n-1}$) converges to 2

The following are some of the important series which are Divergent.

- Infinite sum of all positive integers
- Infinite sum of reciprocal of all positive integers.
- Infinite sum of reciprocal of all prime numbers.

- A sequence can not converge to more than one limit.
- Every convergent sequence is bounded but the converse is not true.

The given sequence <$S_n$> is close to any number as $n \rightarrow \infty$, then we can say that the given series is converge to any number or limit L and denoted by $\sum_{n = 1}^{\infty}a_{n}$ = $\lim_{n\rightarrow \infty }$$S_{n} = L$

We know that the total sum of any series is the limit of the sequence of respective series. So, we say that if $\sum a_{n}$ is the series and the respective sequence is given by <$S_n$>. Then, the given series $\sum a_{n}$ is convergent iff the sequence <$S_n$> is convergent and we can write it as $\lim_{n\rightarrow \infty }$$S_{n}$ = $\sum_{n = 1}^{\infty}a_{n}$

Let us discuss here two most common convergence theorems.

According to this theorem, if $f_{n}$ be given a a sequence of functions that are measurable and defined over a given measurable set M. If $f_{n} \rightarrow f$ is a pointwise function on set M and $f_{n}$ is bounded uniformly bounded, in other words if $|f_{n}(x)|\ \leq \ M$ for all x and n $\in$ N then $lim \int f_{n} = \int f$

This theorem is being widely utilized in probability theory because it provides sufficient condition for convergence of random variables expected value. According to Lebesgue's dominated convergence theorem, let $f_{n}$ be given as a sequence of measurable functions that are real-valued over the measure space (S, $\sum$, m). If this sequence shows pointwise convergence to a function f and is also dominated by another integrable function h in such a way that

$f_{n}(x)| \ \leq h(x)$

for each points x $\in$ S.

Then

$\lim_{n \rightarrow {\infty}} \int_{S} |f_{n} - f| \,dm = 0$

This also gives

$\lim_{n \rightarrow \infty} \int_{S} f_{n} \ dm = \int_{S} f\ dm$

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

For finding this series to be convergent or divergent, we would find if its partial sum converges or diverges.

$lim_{n \rightarrow \infty} \frac{n}{2}(n + 1) = \infty$

Since, the sequence of partial sums diverges

Hence, given series also diverges.

let $S_{n} = \sum_{n=0}^{\infty}(-1)^{n}$

$S_{0} = (-1)^{0}$ = 1

$S_{1} = (-1)^{0} + (-1)^{1}$ = 1 -1 = 0

$S_{2} = (-1)^{0} + (-1)^{1} + (-1)^{2}$ = 1 - 1 + 1 = 1

$S_{3} = (-1)^{0} + (-1)^{1} + (-1)^{2} + (-1)^{3}$ = 1 - 1 + 1 - 1 = 0

and similarly

$S_{4}$ = 1

$S_{5}$ = 0

and so on.

This sequence of partial sums looks like {1, 0, 1, 0, 1, 0, ...} which diverges since $S_{\infty}$ doesn’t exist. Therefore, the series also diverges.

This sequence of partial sums looks like {1, 0, 1, 0, 1, 0, ...} which diverges since $S_{\infty}$ doesn’t exist. Therefore, the series also diverges.

$\sum_{p \geq 0} \frac{4^{2p- 1}}{3^{3p+ 1}}$

If it is convergent, find the value.

**Solution :** $\frac{4^{2p- 1}}{3^{3p+ 1}}$

= $\frac{4^{-1}.4^{2p}}{3.3^{3p}}$

= $\frac{1}{12} \frac{4^{2p}}{3^{3p}}$

= $\frac{1}{12} \frac{16^{p}}{27^{p}}$

= $\frac{1}{12} (\frac{16}{27})^{p}$

Since $\frac{16}{27}$ < 1, we can say that the this partial sum is convergent, hence the given series is convergent too.

Now, we can see that the term $(\frac{16}{27})^{p}$ is a geometric series. So,

$\sum_{p \geq 0} \frac{4^{2p- 1}}{3^{3p+ 1}}$

Since $\frac{16}{27}$ < 1, we can say that the this partial sum is convergent, hence the given series is convergent too.

Now, we can see that the term $(\frac{16}{27})^{p}$ is a geometric series. So,

$\sum_{p \geq 0} \frac{4^{2p- 1}}{3^{3p+ 1}}$

= $\frac{1}{12} \sum_{p \geq 0} (\frac{16}{27})^{p}$

= $\frac{1}{12} \frac{1}{1 - \frac{16}{27}}$

= $\frac{1}{12} \frac{27}{11}$

= $\frac{27}{132}$

Thus, the series converges to $\frac{27}{132}$.

Practice with the following problems based on convergence and divergence.

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