MHB -z.54 find the radius of convergence

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The discussion focuses on finding the radius of convergence for the series given by the formula involving factorials and powers of x. The initial incorrect assumption was that the radius of convergence was 6, but the correct answer is 1/6. The coefficient of x^n simplifies to 6^n, leading to the conclusion that the limit of the nth root of the coefficients is 6. Consequently, the radius of convergence is calculated as the reciprocal of this limit, yielding 1/6. Understanding the simplification of the coefficients is crucial for correctly determining the radius.
karush
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$\tiny{10.7.37}$
$\displaystyle\sum_{n=1}^{\infty}
\frac{6\cdot 12 \cdot 18 \cdots 6n}{n!} x^n$
find the radius of convergence
I put 6 but that wasn't the answer
 
Last edited:
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Please explain how you originally obtained $6$ as the answer.
 
the ans was 1/6
Looked at an example very close to this
and noticed the first term revealed the answer but couldn't follow all the steps they had to get it.
 
Well, the coefficient $a_n$ of $x^n$ in the power series reduces to $6^n$, for $6\cdot 12\cdot 18\cdots 6n = (6\cdot 1)(6\cdot 2)\cdots (6\cdot n) = 6^nn!$. So, $\sqrt[n]{a_n} = 6$, and the radius $R$ of convergence of the power series is given by $1/\lim\limits_{n\to \infty} \sqrt[n]{a_n} = 1/6$.
 

Thread 'Proving that convexity implies second order derivative being positive'

There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...
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