mateomy
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I think I did this right...
<br /> \sum_{i=1}^{\infty} \frac{n}{e^{n^2}}<br />
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...
<br /> \lim_{n \to \infty} \frac{(n+1)}{e^{2n} e(n)}<br />
...which is an indeterminate form (infinity over infinity)
so I take L'Hopital's Rule and come to...
<br /> \frac{1}{2e^{2n} (e)}<br />
Which, when taking the limit goes to zero to show absolute convergence.
Can someone confirm this for me?
Thanks!
<br /> \sum_{i=1}^{\infty} \frac{n}{e^{n^2}}<br />
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...
<br /> \lim_{n \to \infty} \frac{(n+1)}{e^{2n} e(n)}<br />
...which is an indeterminate form (infinity over infinity)
so I take L'Hopital's Rule and come to...
<br /> \frac{1}{2e^{2n} (e)}<br />
Which, when taking the limit goes to zero to show absolute convergence.
Can someone confirm this for me?
Thanks!
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