Ray Vickson said:
So, what did the integral test give you?
Let ##f(x) = \frac {\sqrt x} {e^\sqrt x}##
Note f is a continuous, positive function on ##[1, \infty)##
Note also that ##\frac {d} {dx} f(x) = \frac {1 - \sqrt x} {2e^{\sqrt x} \sqrt x} \leq 0## when ## x \geq 1 \implies## f is a decreasing function.
Let ##u = e^{-\sqrt x} \implies du = \frac {-e^{-\sqrt x}} {2 \sqrt x} dx##
Let ##dv = \sqrt x \, dx \implies v = \frac {2x^{\frac 3 2}} {3}##
Let ##o = -\sqrt x\implies dx = 2o\,do##
Let ##s = o \implies ds = do##
Let ##dt = e^o \, do \implies t = e^o##
Let ##q = x \implies dq = dx##
Let ##dr = e^o \, dx \implies r = 2t(s - 1)##
##\int f \, dx##
##= \int \sqrt x e^{-\sqrt x} dx##
##= \int u \, dv##
##= uv - \int v \, du##
##= uv - \frac 1 3 \int q \, dr##
##= uv - \frac 1 3 (qr - \int r \, dq)##
##= uv - \frac 1 3 (qr - \int 2t(s - 1)) \, dx##
##= uv - \frac 1 3 (qr - 2(\int oe^o \, dx - \int e^o dx))##
##\int \sqrt x e^{-\sqrt x} dx = uv + \frac {qr} 3 + \frac 2 3 \int \sqrt x e^{-\sqrt x} \, dx + \frac 2 3 \int e^{-\sqrt x} \, dx##
##\frac 1 3 \int \sqrt x e^{-\sqrt x} \, dx = uv + \frac {qr} 3 + \frac {2r} 3##
##\int \sqrt x e^{-\sqrt x} \, dx = 2e^{-\sqrt x}x^{\frac 3 2} + 2xt(s - 1) + 4t(s - 1)##
##= 2(e^{-\sqrt x} x^{\frac 3 2} + e^{-\sqrt x}(-\sqrt x - 1)(x + 2))##
##= -2e^{-\sqrt x}(x + 2\sqrt x + 2) + C##
Applying the Integral Test,
##\lim_{t\to\infty} \int_{1}^{t} e^{-\sqrt x} \sqrt x \, dx##
##= \lim_{t\to\infty} -2e^{-\sqrt x}(x + 2\sqrt x + 2)\vert_{1}^{t}##
##= -2(\lim_{t\to\infty} \frac {t + 2\sqrt t + 2} {e^{\sqrt t}} - \frac 5 e)##
##= -2(\lim_{t\to\infty} \frac {1 + \frac 2 {\sqrt t} + \frac 2 t} {\frac {e^{\sqrt t}} t} - \frac 5 e)##
##= -2(\lim_{t\to\infty} \frac 1 {\frac {e^{\sqrt t}} t} - \frac 5 e)##
##= -2(\lim_{t\to\infty} \frac {t} {e^{\sqrt t}} - \frac 5 e)##
##= -2(2\lim_{t\to\infty} \frac 1 {e^{\sqrt t}} - \frac 5 e)##
##= -2(2 \cdot 0 - \frac 5 e)##
##= \frac {10} e \implies## the improper integral is convergent.
##\implies## the given series is convergent by the Integral Test.