Find $\limsup\limits_{j\to\infty} |a_j|^{1/j}$, where $$a_j = \sum_{j=1}^\infty \frac{(1+1/j)^{2j}}{e^j}$$
Since the limit of the numerator is $e^2$, is it correct that the $\limsup$ is equal to $0$? How can I write out the formal calculation?
Asked
Active
Viewed 358 times
1
kiwifruit
- 707
-
1Note that as written, $a_j$ is a constant. – André Nicolas Apr 19 '14 at 02:43
-
I guess this is fine...I wrote the problem exactly as asked – kiwifruit Apr 19 '14 at 02:57
-
Then it is poor notation, but as written, the limsup is $1$. – André Nicolas Apr 19 '14 at 02:58
-
Hmm,but how does that happen? – kiwifruit Apr 19 '14 at 02:59
-
As I noted, $a_j$ is a constant, and for any $a\gt 0$, the limit of $a^{1/j}$ is $1$. – André Nicolas Apr 19 '14 at 03:00
1 Answers
1
Either a typo or a trick question! As written, $a_j$ is constant. (Note that the summation variable $j$ on the right is a "dummy" variable. And the series on the right converges, after a while the terms are close to those of a convergent geometric series.)
For any $a\gt 0$, we have $\lim_{j\to\infty} a^{1/j}=1$.
André Nicolas
- 507,029