Sum Convergent Series

Convergent series are one of the basis for mathematics. Convergent series is defined as the sequencing steps. If the sequences produces in the result are having partial sums, then only the sequence is said to be convergent. The sum of convergent series includes a limit function, point function, line function and value function. Infinite series are also included in the sum convergent series.

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Properties of sum convergent series:

There are many properties for the sum convergent series, They are mentioned below,
1. For example, the series having  ∑ y_n and are having the partial fraction value as P_n+1 , then the formula for this is given by,
P_n+1 = P_n + y_n+1 . here the value of n  `>=` 1.
2. The condition for convergent series for  ∑ y_nis given by the formula,
`lim_(n->oo)` y_n = 0
3. For an infinite series, the sum for convergent series is given by,
`sum_(n=0)^(n=oo)` pⁿ =1 + p + p² +....... = `(1)/(1-p)`

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Example problem for sum convergent series:

Question 1: Find the convergent series and the sum for the given problem,  `sum_(n>=1)` `(4^n+2^n)/(6^n)` .
Solution:
Step 1: Finding the convergent series by using the condition, we get,
`(4^n + 2^n)/(6^n)`  = `(4^n)/(6^n)`  + `(2^n)/(6^n)`  = `((4)/(6))^n` + `((2)/(6))^n`
Step 2: By using the properties of sum convergent series, we are going to find the sum,
`sum_(n=0)^(n=oo)` `(4^n + 2^n)/(6^n)`  = `sum_(n=0)^(n=oo)` `((4)/(6))^n` + `((2)/(6))^n`
Step 3: Then by simplification, we get,
              = `(1)/(1-(4/6))` +`(1)/(1-(2/6))`  = `(16)/(4)`  = 4
This is the required sum for the convergent series.

Question 2: Find the convergent series and the sum for the given problem,  `sum_(n>=1)` `(2^n+2^n)/(6^n)` .
Solution:
Step 1: Finding the convergent series by using the condition, we get,
`(2^n + 2^n)/(6^n)`  = `(2^n)/(6^n)`  + `(2^n)/(6^n)`  = `((2)/(6))^n` + `((2)/(6))^n`
Step 2: By using the properties of sum convergent series, we are going to find the sum,
`sum_(n=0)^(n=oo)` `(2^n + 2^n)/(6^n)`  = `sum_(n=0)^(n=oo)` `((2)/(6))^n` + `((2)/(6))^n`
Step 3: Then by simplification, we get,
              = `(1)/(1-(2/6))` +`(1)/(1-(2/6))`  = `(6)/(2)`  =3
This is the required sum for the convergent series.