Test the following series for convergence or divergence

In summary: R}## Edit: The next section is on power series. So they don't define e in terms of power series? Do they define the power series for ##e^x## and prove it converges for all ##x##?Edit: The next section is on power series. So they don't define e in terms of power series? Do they define the power series for
  • #1
Entertainment Unit
16
1

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.
 
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  • #2
Entertainment Unit said:

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.
 
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  • #3
Entertainment Unit said:

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?
 
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  • #4
Ray Vickson said:
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.
 
  • #5
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.
 
  • #6
hmmmm,
Entertainment Unit said:
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
 
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  • #7
StoneTemplePython said:
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!
 
  • #8
Entertainment Unit said:
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 ##
 

Related to Test the following series for convergence or divergence

1. What is the purpose of testing a series for convergence or divergence?

The purpose of testing a series for convergence or divergence is to determine whether the series will have a finite or infinite sum. This is important in mathematical and scientific calculations, as it helps to determine the behavior of the series and its overall value.

2. How do you test a series for convergence or divergence?

To test a series for convergence or divergence, one can use various methods such as the ratio test, the root test, or the integral test. These methods involve evaluating the limit of certain ratios or integrals to determine if the series converges or diverges.

3. What is the difference between a convergent and a divergent series?

A convergent series is one that has a finite sum, meaning that the terms of the series approach a specific value as the number of terms increases. A divergent series, on the other hand, has an infinite sum, meaning that the terms of the series do not approach a specific value and the series does not have a finite sum.

4. What are some common types of series that are frequently tested for convergence or divergence?

Some common types of series that are frequently tested for convergence or divergence include geometric series, p-series, alternating series, and Taylor series. These types of series have specific rules and tests that can be used to determine their convergence or divergence.

5. What are some real-world applications of testing series for convergence or divergence?

Testing series for convergence or divergence has many real-world applications, such as in finance, physics, and engineering. In finance, it can be used to calculate compound interest or determine the value of investments. In physics, it can be used to model the behavior of physical systems. In engineering, it can be used to analyze the stability and performance of systems.

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