- #1
mateomy
- 307
- 0
I think I did this right...
[tex]
\sum_{i=1}^{\infty} \frac{n}{e^{n^2}}
[/tex]
I tried it with the root test to no avail. So I then tried it with the Ratio Test and I come to this expression...
[tex]
\lim_{n \to \infty} \frac{(n+1)}{e^{2n} e(n)}
[/tex]
...which is an indeterminate form (infinity over infinity)
so I take L'Hopital's Rule and come to...
[tex]
\frac{1}{2e^{2n} (e)}
[/tex]
Which, when taking the limit goes to zero to show absolute convergence.
Can someone confirm this for me?
Thanks!
[tex]
\sum_{i=1}^{\infty} \frac{n}{e^{n^2}}
[/tex]
I tried it with the root test to no avail. So I then tried it with the Ratio Test and I come to this expression...
[tex]
\lim_{n \to \infty} \frac{(n+1)}{e^{2n} e(n)}
[/tex]
...which is an indeterminate form (infinity over infinity)
so I take L'Hopital's Rule and come to...
[tex]
\frac{1}{2e^{2n} (e)}
[/tex]
Which, when taking the limit goes to zero to show absolute convergence.
Can someone confirm this for me?
Thanks!
Last edited: