Answer
Convergent$\;,\; 1-\large\frac{1}{e}$
Work Step by Step
Let
\[I=\int_{1}^{\infty}\frac{e^{\frac{-1}{x}}}{x^2}dx\]
\[I=\lim_{t\rightarrow \infty}\int_{1}^{t}\frac{e^{\frac{-1}{x}}}{x^2}dx\;\;\;\ldots(1)\]
\[I=\lim_{t\rightarrow \infty}\left[e^{-\frac{1}{x}}\right]_{1}^{t}\]
\[I=\lim_{t\rightarrow \infty}\left[e^{-\frac{1}{t}}-e^{-1}\right]\]
\[I=1-\frac{1}{e}\]
Since limit on R.H.S. of (1) exists so $I$ is convergent
and $I=1-\large\frac{1}{e}$.
