Test the following series for convergence or divergence

[Entertainment Unit]

Homework Statement

Test the following series for convergence or divergence.

$\sum_{n = 1}^{\infty} \frac {\sqrt n} {e^\sqrt n}$

Homework Equations

None that I'm aware of.

The Attempt at a Solution

I know I can use the Integral Test for this, but I was hoping for a simpler way.

 

[member 587159]

Homework Statement

Test the following series for convergence or divergence.

$\sum_{n = 1}^{\infty} \frac {\sqrt n} {e^\sqrt n}$

Homework Equations

None that I'm aware of.

The Attempt at a Solution

I know I can use the Integral Test for this, but I was hoping for a simpler way.

Intuitively, it should converge because of the exponential in the denominator. I think integral test is fine here. I tried d'Alembert's criterion but unless I made a mistake by being too quick the limit gives 1 so the test is inclusive.

I am still looking for something to compare the series with.

 

[Ray Vickson]

Homework Statement

Test the following series for convergence or divergence.

$\sum_{n = 1}^{\infty} \frac {\sqrt n} {e^\sqrt n}$

Homework Equations

None that I'm aware of.

The Attempt at a Solution

I know I can use the Integral Test for this, but I was hoping for a simpler way.

So, what did the integral test give you?

 

[Entertainment Unit]

So, what did the integral test give you?

Point taken, I need to get over my laziness and just calculate the integral. I was hoping for one of those "why couldn't / didn't I think of that" moments in the form of a series to use to compare the given series with.

 

[Entertainment Unit]

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.

 

[StoneTemplePython]

hmmmm,

I know I can use the Integral Test for this, but I was hoping for a simpler way.

my vote we be to note that for $x \gt 0$, we have
$0\lt \frac{1}{6!}x^6 \leq e^x$
justification: look at power series

inverting, then
$\frac{1}{e^x} \leq 6! \frac{1}{x^6}$

selecting $x := \sqrt{n}$ and rescaling each side of the inequality by $\sqrt{n}$
there's an immediate pointwise bound, for $n \geq 2$

which you can bound above by something very nice
(hint: make use of $n\geq 2$ for something nice)

then sum over the bound

 

[Entertainment Unit]

hmmmm, ...

Thank you, the next section in my textbook is entitled "Power Series" so once I'm done this set of exercises, I'll work through that and come back to your post. Thanks again!

 

[StoneTemplePython]

Thank you, the next section in my textbook is entitled "Power Series" so once I'm done this set of exercises, I'll work through that and come back to your post. Thanks again!

oh ok -- out of curiosity, what do they define $e$ or the exponential function as? (Maybe just a black box right now?)

the something nice I had mind, by the way is that

$\sum_{j=2}^m \frac{1}{j(j-1)} = 1 - \frac{1}{m}$
this is a very nice telescoping sum and a good little exercise if you have the time. Maybe you've already seen it before in class.

so
$\lim_{m\to \infty}\sum_{j=2}^m \frac{1}{j(j-1)}= \sum_{j=2}^\infty \frac{1}{j(j-1)} = 1$