Let $0
To prove divergence of a series.
4
$\begingroup$
calculus
-
3If $N$ is fixed, what is $\lim\limits_{r\rightarrow1^-}\sum_{j=1}^N r^{j!}$? – 2012-12-16
1 Answers
1
Let $N$ be arbitrary but fixed; to show divergence of $\sum_n r^{n!}$ as $r \to 1$ we need to be able to find an $r$ such that $\sum_n r^{n!} > N$.
As the comment by David Mitra hints at, we can already arrange for the first $N+1$ terms in the summation to exceed $N$. For, take:$r = \sqrt[(N+1)!]{\frac N{N+1}}$
Then it is not too hard to see that if $n \le N+1$, the $n$th summand exceeds $\dfrac N{N+1}$ and therefore that the entire sum must exceed $N$.
The same argument works, mutatis mutandis, for other powers than $n!$.