Answer:
Yes
Step-by-step explanation:
[tex]\sum_{n=1}^{\infty} e^{-\sqrt{n} } \\\sum_{n=1}^{\infty} (\frac{1}{e})^{\sqrt{n} }[/tex]
If we say u = √n, then:
[tex]\sum_{u=1}^{\infty} (\frac{1}{e})^{u}[/tex]
This is a geometric series. Since |1/e| < 1, the series converges.
