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Question: How do you test the series \[\sum {\dfrac{1}{{n!}}} \] from \[n\] is \[\left[ {0,\infty } \right)\] ...

How do you test the series ∑1n!\sum {\dfrac{1}{{n!}}} from nn is [0,∞)\left[ {0,\infty } \right) for convergence?

Explanation

Solution

Here, we will apply the ratio test on the given series. Then we will solve the limit in the test and find the answer. According to the answer, we will be able to know whether the given series is convergent, divergent, or inconclusive.

Formula Used:
Ratio test: L=lim⁡n→∞∣an+1an∣=rL = \mathop {\lim }\limits_{n \to \infty } \left| {\dfrac{{{a_n} + 1}}{{{a_n}}}} \right| = r.

Complete step-by-step answer:
Given series is ∑1n!\sum {\dfrac{1}{{n!}}}
Now, in order to test that the given series ∑1n!\sum {\dfrac{1}{{n!}}} from nn is [0,∞)\left[ {0,\infty } \right) for convergence, we will use the ratio test.
According to the ratio test, It says that there exists rr such that for the series ∑an\sum {{a_n}} we can make a determination about its convergence by taking
L=lim⁡n→∞∣an+1an∣=rL = \mathop {\lim }\limits_{n \to \infty } \left| {\dfrac{{{a_n} + 1}}{{{a_n}}}} \right| = r
Here, if r<1r < 1, then the series ∑an\sum {{a_n}} is absolutely convergent.
If r>1r > 1, then the series ∑an\sum {{a_n}} is divergent.
And, if r=1r = 1, then the ratio test is inconclusive, and the series may converge or diverge.
Hence, for the series ∑n=0∞\sum\nolimits_{n = 0}^\infty {} we let an=1n!{a_n} = \dfrac{1}{{n!}}
Then, we get,
L=lim⁡n→∞∣1(n+1)!1n!∣=lim⁡n→∞∣n!(n+1)!∣L = \mathop {\lim }\limits_{n \to \infty } \left| {\dfrac{{\dfrac{1}{{\left( {n + 1} \right)!}}}}{{\dfrac{1}{{n!}}}}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {\dfrac{{n!}}{{\left( {n + 1} \right)!}}} \right|
We know that, (n+1)!\left( {n + 1} \right)! can be written as: (n+1)×n!\left( {n + 1} \right) \times n!
⇒lim⁡n→∞∣n!(n+1)!∣=lim⁡n→∞∣n!(n+1)×n!∣=lim⁡n→∞∣1(n+1)∣=0\Rightarrow \mathop {\lim }\limits_{n \to \infty } \left| {\dfrac{{n!}}{{\left( {n + 1} \right)!}}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {\dfrac{{n!}}{{\left( {n + 1} \right) \times n!}}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {\dfrac{1}{{\left( {n + 1} \right)}}} \right| = 0
Since, L=r=0L = r = 0
Therefore, this means that r<1r < 1
Therefore, ∑an=∑n=0∞1n!\sum {{a_n}} = \sum\limits_{n = 0}^\infty {\dfrac{1}{{n!}}} is convergent through [0,∞)\left[ {0,\infty } \right).
Hence, we have tested the series ∑1n!\sum {\dfrac{1}{{n!}}} from nn is [0,∞)\left[ {0,\infty } \right) for convergence
Thus, this is the required answer.

Note: In mathematics, convergence is a property which is exhibited by certain infinite series and functions of approaching a limit more and more closely as an argument or variable of the function increases or decreases or as the number of terms of the series increases. The convergent sequence is when through some terms we achieve a final and constant term as nn approaches infinity. The divergent sequence is that in which the terms never become constant they continue to increase or decrease and they approach infinity or minus infinity as nn approaches infinity.