You've established that
[tex]\displaystyle\sum_{n\ge1}\frac1{2^n}\le\int_1^\infty\frac{\mathrm dx}{2^x}[/tex]
so all you need to do is compute the integral. Rewrite the integrand as
[tex]\dfrac1{2^x}=2^{-x}=e^{\ln2^{-x}}=e^{-x\ln2}[/tex]
and replace [tex]y=-x\ln2[/tex], so that [tex]\mathrm dy=-\ln2\,\mathrm dx[/tex]. Then
[tex]\displaystyle\int_{x=1}^{x\to\infty}\frac{\mathrm dx}{2^x}=\int_{y=-\ln2}^{y\to-\infty}e^y\,\frac{\mathrm dy}{-\ln 2}=\frac1{\ln2}\int_{-\infty}^{-\ln2}e^y\,\mathrm dy[/tex]
By the fundamental theorem of calculus, this evaluates to
[tex]\dfrac1{\ln2}e^y\bigg|_{y\to-\infty}^{y=-\ln2}=\dfrac1{\ln2}\left(e^{-\ln2}-\lim_{y\to-\infty}e^y\right)=\dfrac{e^{-\ln2}}{\ln2}=\dfrac{\frac12}{\ln2}=-\dfrac1{2\ln2}=\dfrac1{\ln4}[/tex]
and so the series converges.
You also could have used the fact that the series is geometric with common ratio less than 1 to arrive at the same conclusion, with the added perk of being able to find the exact value of the sum to corroborate this.
