I came across the following power series while looking at a problem.

$$-1+a_1x+a_2x^2+a_3x^3+.......$$ where $a_0=-1$ and the choice of $a_1$ is arbitrary.The other coefficients are dependent on $a_1$ in the following way : $$a_n= (-1)^{(n+1)} \frac{n^{n-1}}{(n-1)!}.a_1^{n}$$

For $a_1=0$,such a power series clearly converges.Can we find other values of $a_1$ such that the above power series converges for all $x$. To move ahead,I was considering this:

If $|a_n| \leq \frac{c}{n!}$ where $c$ is a fixed constant,then the series converges.Can we find such a constant $c$ for some value of $a_1$ other than 0.?

  • $\begingroup$ $\le \frac{c}{n!}$ is I think not achievable. But for suitable values of $a_1$ look at the Stirling approximation to $n!$. $\endgroup$ – André Nicolas Nov 30 '15 at 1:09

$|{a_{n+1} \over a_n}| = { (n+1)^n a_1^{n+1} \over n!} {(n-1)! \over n^{n-1} a_1^n } = (1+ {1 \over n})^n a_1$ and $\lim_n (1+ {1 \over n})^n a_1 = e a_1$.

Hence the radius of convergence is $R = {1 \over e a_1}$. In order that the power series is entire, you need $a_1 = 0$.


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