I was trying to get the radius of convergence for the following question but get stuck along the way.

$\sum_{n=0}^\infty \frac{(n!)^k}{(kn)!}x^n$

By using ratio test:

$lim_{n\rightarrow\infty}|{\frac{(n+1)!^k}{(k(n+1)) !}x^{n+1}/\frac{(n!)^k}{(kn)!}x^n}| <1$

but i can't seem to simplify $\frac{(kn)!}{k(n+1)!}$

The answer is: $k^k$

2. Originally Posted by noob mathematician
I was trying to get the radius of convergence for the following question but get stuck along the way.

$\sum_{n=0}^\infty \frac{(n!)^k}{(kn)!}x^n$

By using ratio test:

$lim_{n\rightarrow\infty}|{\frac{(n+1)!^k}{(k(n+1)) !}x^{n+1}/\frac{(n!)^k}{(kn)!}x^n}| <1$

but i can't seem to simplify $\frac{(kn)!}{k(n+1)!}$

The answer is: $k^k$

Sure that is the answer. You began correctly but stopped short of opening and simplifying stuff a little bit more:

$\frac{(kn)!}{(k(n+1))!}=\frac{(kn)!}{(kn+k)!}=\fra c{1}{(kn+1)(kn+2)\cdot...\cdot(kn+k)}$ $=\frac{1}{k^kn^k+\{\mbox{terms with n to a power less than k }\}}$

Now go back to your limit and plug in the above and you'll get the correct answer.

Tonio

3. Originally Posted by noob mathematician
I was trying to get the radius of convergence for the following question but get stuck along the way.

$\sum_{n=0}^\infty \frac{(n!)^k}{(kn)!}x^n$

By using ratio test:

$lim_{n\rightarrow\infty}|{\frac{(n+1)!^k}{(k(n+1)) !}x^{n+1}/\frac{(n!)^k}{(kn)!}x^n}| <1$

but i can't seem to simplify $\frac{(kn)!}{k(n+1)!}$

The answer is: $k^k$
$a_n = \frac{{{{\left( {n!} \right)}^k}}}{{\left( {kn} \right)!}}{x^n} \Rightarrow {a_{n + 1}} = \frac{{{{\left( {\left( {n + 1} \right)!} \right)}^k}}}{{\left( {kn + k} \right)!}}{x^{n + 1}} = \frac{{{{\left( {n!} \right)}^k}{{\left( {n + 1} \right)}^k}}}{{\left( {kn} \right)!\left( {kn + 1} \right)\left( {kn + 2} \right) \ldots \left( {kn + k} \right)}}{x^{n + 1}}.$

$\frac{{{a_{n + 1}}}}{{{a_n}}} = \frac{{x{{\left( {n + 1} \right)}^k}}}{{\left( {kn + 1} \right)\left( {kn + 2} \right) \ldots \left( {kn + k} \right)}} = \frac{{x{{\left( {n + 1} \right)}^k}}}{{\prod\limits_{i = 1}^k {\left( {kn + i} \right)} }}.$

$\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{a_{n + 1}}}}
{{{a_n}}}} \right| = \left| x \right|\mathop {\lim }\limits_{n \to \infty } \frac{{{{\left( {n + 1} \right)}^k}}}{{\prod\limits_{i = 1}^k {\left( {kn + i} \right)} }} = \left| x \right|\mathop {\lim }\limits_{n \to \infty } \frac{{{n^k}{{\left( {1 + \frac{1}{n}} \right)}^k}}}{{{k^k}{n^k}\prod\limits_{i = 1}^k {\left( {1 + \frac{i}{{kn}}} \right)} }} =$

$= \frac{{\left| x \right|}}
{{{k^k}}}\mathop {\lim }\limits_{n \to \infty } \frac{{{{\left( {1 + \frac{1}
{n}} \right)}^k}}}
{{\prod\limits_{i = 1}^k {\left( {1 + \frac{i}
{{kn}}} \right)} }} = \frac{{\left| x \right|}}
{{{k^k}}} \cdot \frac{{{{\left( {1 + 0} \right)}^k}}}
{{\prod\limits_{i = 1}^k {\left( {1 + 0} \right)} }} = \frac{{\left| x \right|}}
{{{k^k}}}.$

$\frac{{\left| x \right|}}{{{k^k}}} < 1 \Leftrightarrow - {k^k} < x < {k^k} \Rightarrow R = k^k.$