Question

What is the interval of convergence of $\sum\limits_1^\infty {\dfrac{{{{( - x)}^{2n + 1}}}}{{(2n + 1)!}}}$?

Answer 1

To solve the problem we will apply the Ratio test.
Let us consider, the $\mathop {\lim }\limits_{n \to \infty} \left| {\dfrac{{{a_{n + 1}}}}{{{a_n}}}} \right| = L$
When $L < 1$, then $\sum\limits_1^0 {{a_n}}$ converges.
When $L > 1$, then $\sum\limits_1^0 {{a_n}}$ diverges.
When$L = 1$,then the test is inconclusive.

Complete step-by-step solution:
It is given that; the series is $\sum\limits_1^\infty {\dfrac{{{{( - x)}^{2n + 1}}}}{{(2n + 1)!}}}$.
We have to find the interval of convergence of $\sum\limits_1^\infty {\dfrac{{{{( - x)}^{2n + 1}}}}{{(2n + 1)!}}}$.
To find the interval we will apply a ratio test.
Let us consider, the $\mathop {\lim }\limits_{n \to 0} \left| {\dfrac{{{a_{n + 1}}}}{{{a_n}}}} \right| = L$
When $L < 1$, then $\sum\limits_1^0 {{a_n}}$ converges.
When $L > 1$, then $\sum\limits_1^0 {{a_n}}$ diverges.
When$L = 1$, then the test is inconclusive.
Now,
$(\left| {\dfrac{{{a_{n + 1}}}}{{{a_n}}}} \right|) = \left( {\left| {\dfrac{{\dfrac{{{{( - x)}^{2(n + 1) + 1}}}}{{(2(n + 1) + 1)!}}}}{{\dfrac{{{{( - x)}^{2n + 1}}}}{{(2n + 1)!}}}}} \right|} \right)$
Now, we calculate the limits.
$\mathop {\lim }\limits_{n \to \infty } \left( {\left| {\dfrac{{\dfrac{{{{( - x)}^{2(n + 1) + 1}}}}{{(2(n + 1) + 1)!}}}}{{\dfrac{{{{( - x)}^{2n + 1}}}}{{(2n + 1)!}}}}} \right|} \right)$
Simplifying we get,
$= \mathop {\lim }\limits_{n \to \infty } \left( {\left| {\dfrac{{\dfrac{{{{( - x)}^{2n + 3}}}}{{(2n + 3)!}}}}{{\dfrac{{{{( - x)}^{2n + 1}}}}{{(2n + 1)!}}}}} \right|} \right)$
Simplifying again we get,
$= \mathop {\lim }\limits_{n \to \infty } \left( {\left| {\dfrac{{{{( - x)}^{2n + 3}}}}{{{{( - x)}^{2n + 1}}}} \times \dfrac{{(2n + 1)!}}{{(2n + 3)!}}} \right|} \right)$
Simplifying again we get,
$= \mathop {\lim }\limits_{n \to \infty } \left( {\left| {{x^2} \times \dfrac{1}{{(2n + 3)(2n + 2)}}} \right|} \right)$
Simplifying again we get,
$= \mathop {\lim }\limits_{n \to \infty } \left( {\left| {{x^2} \times \dfrac{1}{{(2n + 3)2(n + 1)}}} \right|} \right)$
Simplifying again we get,
$= \left| {\dfrac{{{x^2}}}{2}} \right|\mathop {\lim }\limits_{n \to \infty } \left( {\left| {\dfrac{1}{{(2n + 3)(2n + 2)}}} \right|} \right)$
Simplifying again we get,
$= \left| {\dfrac{{{x^2}}}{2}} \right| \times 0$
Simplifying again we get,
$= 0$
Therefore, $L < 1$ for every $x$
So, $\sum\limits_1^\infty {\dfrac{{{{( - x)}^{2n + 1}}}}{{(2n + 1)!}}}$ converges for all $x$.
Hence, the interval is $- \infty < x < \infty$

Note: Generally, the ratio test is a test for the convergence of a series $\sum\limits_1^\infty {{a_n}}$ where each term is a real or complex number and an is nonzero when n is large.
The ratio test states that:

When L < 1, then the series converges absolutely;
When L > 1 then the series is divergent;
When L = 1 or the limit fails to exist, then the test is inconclusive, because there exist both convergent and divergent series that satisfy this case.
It is possible to make the ratio test applicable to certain cases where the limit L fails to exist, if limit superior and limit inferior are used. The test criteria can also be refined so that the test is sometimes conclusive even when L = 1.