Question

How do you find the sum of an infinite non-geometric series?

Answer 1

Hint: There is no definite way to find the sum of an infinite non-geometric series. It is quite difficult to find the sum of an infinite non-geometric series , you do it by the definition of sum of a series; i.e., partial sums. You then consider whether these numbers sum are becoming closer and closer to at least one number. Of course, the mathematical term for this is often taking the limit of ${S_n}$ as $n$ approaches infinity.

Complete step by step solution:
Let us consider a series ,
$\dfrac{1}{2} + \dfrac{1}{6} + \dfrac{1}{{12}} + \dfrac{1}{{20}} + ......\dfrac{1}{{n(n + 1)}} + ....$
This is an infinite non-geometric series. It is quite difficult to find the sum of an infinite non-geometric series , you do it by the definition of sum of a series; i.e., partial sums.
Therefore, the partial sum is given as ,
\[\sum\limits_{n = 1}^N {\dfrac{1}{{n(n + 1)}} = } \sum\limits_{n = 1}^N {\dfrac{{(n + 1) - n}}{{n(n + 1)}}} \] (partial sum up to $Nth$ term)
$ = \sum\limits_{n = 1}^N {\left( {\dfrac{1}{n} - \dfrac{1}{{n + 1}}} \right)} $
 $
   = \sum\limits_{n = 1}^N {\dfrac{1}{n} - \sum\limits_{n = 2}^{N + 1} {\dfrac{1}{n}} } \\
   = 1 + \sum\limits_{i = 2}^N {\dfrac{1}{n} - } \sum\limits_{n = 2}^N {\dfrac{1}{n}} - \dfrac{1}{{N + 1}} \\
   = 1 - \dfrac{1}{{N + 1}} \\
 $
Now for infinite series , if sum is becoming closer and closer to a definite number. Of course, the mathematical term for this is often taking the limit of ${S_n}$ as $n$ approaches infinity.
Sum of infinite non-geometric series is given by ,
\[\mathop {\lim }\limits_{N \to \infty } \sum\limits_{n = 1}^N {\dfrac{1}{{n(n + 1)}} = } \mathop {\lim }\limits_{N \to \infty } \left( {1 - \dfrac{1}{{N + 1}}} \right)\]
                              $ = 1$
We get the required result.

Note: If sum are becoming closer and closer to at least one number. Of course, the mathematical term for this is often taking the limit of ${S_n}$ as $n$ approaches infinity , then we will say that the series converges (to the limit number) and call that answer the sum. If not, we will say that the series diverges.