Question

Test whether the series ∞Σn=0 1/(n^5+x^3) converges uniformly or not

Answer 1

Solution

For the given series

$\sum_{0}^{\infty }\frac{1}{n^5+x^3}$

We use the Ratio Test for the Radius of convergence

Convergence when $L < 1$

$L=\lim_{n\rightarrow \infty }\left | \frac{a_{n+1}}{a_{n}} \right |$

Here, $a_{n}=\frac{1}{n^5+x^3}$ , then

$a_{n+1}=\frac{1}{(n+1)^5+x^3}$

Therefore,

$L=\lim_{n\rightarrow \infty }\left | \frac{\frac{1}{(n+1)^5+x^3}}{\frac{1}{n^5+x^3}}\right |$

$L=\lim_{n\rightarrow \infty }\left | \frac{n^5+x^3}{(1+n)^5+x^3}\right |$

L = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{n^5}\left( {1 + \frac{{{x^3}}}{{{n^5}}}} \right)}}{{{n^5}{{\left( {\frac{1}{n} + 1} \right)}^5} + {x^3}}}} \right|\

L = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{n^5}\left( {1 + \frac{{{x^3}}}{{{n^5}}}} \right)}}{{{n^5}\left( {{{\left( {\frac{1}{n} + 1} \right)}^5} + \frac{{{x^3}}}{{{n^5}}}} \right)}}} \right|\

L = \left| {\frac{{\left( {1 + \frac{{{x^3}}}{{{\infty ^5}}}} \right)}}{{\left( {{{\left( {\frac{1}{\infty } + 1} \right)}^5} + \frac{{{x^3}}}{{{\infty ^5}}}} \right)}}} \right|\

L = \left| {\frac{{\left( {1 + 0} \right)}}{{\left( {{{\left( {0 + 1} \right)}^5} + 0} \right)}}} \right|\

$L=1$

Hence the series may be divergent, conditionally convergent, or absolutely convergent.

Now, when

\sum\limits_0^\infty {\left| {{a_n}} \right|} \ converges, then \sum\limits_0^\infty {{a_n}} \ converges.

Therefore, the series is convergent for $x>-1$