Absolute convergence of a series means series will converge even when we take the absolute value of each term. An infinite sequence of numbers is called a series. A series is said to either converge or diverge. When we take the partial sum of an infinite series be it $S_1,\ S_2,\ S_3,$ … and if the partial sum has a finite value or tends to a limit, then the series is said to be convergent. $\sum_{k = 1}^n$ $s_k$

Convergence of an infinite series could be of two types – absolutely convergent and conditionally convergent. An infinite series suppose $\sum a_n$ is termed absolutely convergent if the summation of each of the absolute value of the term of the series is converging. That is if $\sum a_n$ is converging and $\sum a_n$ is also converging, then the series is called absolutely convergent. It is to be remembered that if a series $\sum a_n$ is absolutely convergent then the series $\sum a_n$ an is also convergent but the vice versa is not true. Which mean, if a series $\sum a_n$ an is convergent its not necessary that the series $\sum a_n$ would also be absolutely convergent.
Absolute convergence of a series is related to convergence of that series such that if the summation of the terms taken partially is tending towards a real value and is finite then the infinite sequence is said to be converging and if summation of the absolute value of the terms which mean the partial sum of all positive value of the terms are taken and it is finite in nature then it is said to be absolutely convergent. There is a fact related to absolute convergence of an infinite sequence termed as absolute convergence test. The absolute convergence test can be stated as if $\sum u_n$ is convergent, then $\sum u_n$ is also convergent but the reverse of the statement does not hold good. That is if we take partial sum of the terms of a series considering them to be all positive and the series converges (called absolutely convergent) then the summation of the series would automatically and definitely be convergent but the converse is not true.
Let us take up the function $f(x)$ having lower limit of integral as $1$ and upper limit of integral as infinity. Then the definite integral would look like
$\int_{x = 1}^{\infty}$ $f(x)\ dx$
If this integral having infinite number of terms converges to a finite value and on taking the integral of absolute value of each term of the same function having same upper and the lower limit as $1$ to infinity also converges to a real value then that type of convergence is called absolutely convergent. There exists an absolute convergent test for an improper integral which states that if an integration of absolute value of a function converges, then the integration of that function also converges. Converse of absolute convergence test statement is not true. That is if we carry out the integration of a function and it converges then the integration of absolute value of the same function shall not converge and if that is the case then it is called conditionally convergent instead of absolutely convergent.
Example 1 :
Test the convergence of the following series:
$\sum_{n = 1}^{\infty}$ $(\frac{1}{4})^{n + 1}$
Solution:
The given problem is geometric series where the number of terms starts from one and moves on to infinity. The common ration of the given geometric series is $r$ = $\frac{1}{4}$. As the value of the common ration is $\frac{1}{4}$ which is less than $1$ so the series converges. Now, let us take up the absolute value of each of these terms.
$\sum_{n = 1}^{\infty}$ $(\frac{1}{4})^{n + 1}$
As we are considering the absolute value of each of the terms it is the summation of the positive values of the series, we disregard the negative sign of any term if present. That series too is convergent. So, as both the summation of the series and the summation of absolute value of each term of the series converges, we can say that the series is absolutely convergent.
Example 2:
$\sum_{n = 1}^{\infty}$ $sin$ $\frac{(\frac{\pi}{2}\ +\ n \times \pi)}{n^{\frac{7}{2}}}$
When $n$ = $2,\ sin$ $(\frac{\pi}{2}$ + $2 \times \pi)$ = $sin$ $\frac{5 \pi}{2}$ = $1$
When $n$ = $3,\ sin$ $(\frac{\pi}{2}$ + $3 \times \pi)$ = $sin$ $\frac{7 \pi}{2}$ = $1$
When $n$ = $4,\ sin$ $(\frac{\pi}{2}$ + $4 \times \pi)$ = $sin$ $\frac{9 \pi}{2}$ = $1$
$\sum_{n = 1}^{\infty}$ $\frac{(\frac{\pi}{2} + n \times \pi)}{n^\frac{7}{2}}$
Test the convergence of the above given series
Solution:
We need to consider the numerator $sin$ ($\frac{\pi}{2}$ + $n \times \pi$) first. So we plugin few values of n starting from $n$ =$1$ to understand the nature of the series.
When $n$ = $1,\ sin$ $(\frac{\pi}{2}$ + $1 \times \pi)$ = $sin$ $\frac{3 \pi}{2}$ = $1$
When $n$ = $2,\ sin$ $(\frac{\pi}{2}$ + $2 \times \pi)$ = $sin$ $\frac{5 \pi}{2}$ = $1$
When $n$ = $3,\ sin$ $(\frac{\pi}{2}$ + $3 \times \pi)$ = $sin$ $\frac{7 \pi}{2}$ = $1$
When $n$ = $4,\ sin$ $(\frac{\pi}{2}$ + $4 \times \pi)$ = $sin$ $\frac{9 \pi}{2}$ = $1$
We see that the terms are alternately negative and positive so we term it as an alternating series. Therefore, the series could be written as
$\sum_{n = 1}^{\infty}$ $\frac{(1)^n}{n^{\frac{7}{2}}}$
To test the infinite sequence whether it is absolutely convergent or not we take the absolute value of the series
$\sum_{n = 1}^{\infty}$ $\frac{(1)^n}{n^{\frac{7}{2}}}$ = $\sum_{n = 1}^{\infty}$ $\frac{1}{n^{\frac{7}{2}}}$
The absolute value of the series is equivalent to summation of series from $n$ = $1$ to infinity of $\frac{1}{n^{\frac{7}{2}}}$. On conducting absolute convergent test that is if the summation of the series of absolute value of each term converges then the series would also converge we can decide if the given series is absolutely convergent or not. So, applying $p$ – series test, the value of $p$ = $\frac{7}{2}$ which is greater than $1$. As the value of $p$ is greater than $1$ so we can say that summation of series from $n$ = $1$ to infinity of $\frac{1}{n^{\frac{7}{2}}}$ is convergent and thus the given series is absolutely convergent in nature.
Example 3:
Test the convergence of the following series:
$\sum_{n = 1}^{\infty}$ $(\frac{1}{2})^{n + 1}$
Solution:
The given problem is geometric series. The common ration of the given geometric series is $r$ = $\frac{1}{2}$. As the value of the common ratio is $\frac{1}{2}$ which is more than $1$ so the series diverges. Therefore, the given series is divergent