- #1
middleCmusic
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Note: This is not strictly a homework problem. I'm just doing these problems for review (college is out for the semester) - but I wasn't sure if putting them on the main part of the forum would be appropriate since they are clearly lower-level problems.(Newbie)
The problem says to "Determine whether the series is convergent or divergent [using only the Comparison Test or the Integral Test]."
The first, #13, is [itex]\sum_{n=1}^{\infty }\, n e^{-n^{2}}[/itex]
and the second, #14, is [itex]\sum_{n=1}^{\infty }\frac{\ln n}{n^{2}}[/itex].
The Integral Test listed in Stewart says:
Suppose [itex]f[/itex] is a continuous, positive, decreasing function on [0,∞) and let [itex]a_n=f(n)[/itex]. Then the series [itex]\sum_{n=1}^{\infty }\[\itex] is convergent if and only if the improper integral [itex]\int_{1}^{\infty }f(x)dx[\itex] is convergent. In other words:
(a) If [itex] \int_{1}^{\infty }f(x)dx [/itex] is convergent, then [itex] \sum_{n=1}^{\infty }a_n [/itex] is convergent.
(b) If [itex] \int_{1}^{\infty }f(x)dx [/itex] is divergent, then [itex] \sum_{n=1}^{\infty}a_n [/itex] is divergent.
The Comparison Test listed in Stewart says
Suppose that [itex]\sum a_n[/itex] and [itex]\sum b_n[/itex] are series with positive terms.
(a) If [itex]\sum b_n[/itex] is convergent and [itex]a_n \leq b_n[/itex] for all [itex]n[/itex], then [itex]\sum a_n[/itex] is also convergent.
(b) If [itex]\sum b_n[/itex] is divergent and [itex]a_n \geq b_n[/itex] for all [itex]n[/itex], then [itex]\sum a_n[/itex] is also divergent.
First, I checked to see if either sequence beings summed had a limit as [itex]n \rightarrow \infty [/itex] that was nonzero, as this would show them to be divergent, but both sequences do go to zero in the limit.
Next, I tried evaluating #13 as a function of x: [itex]f(x)=x e^{-x^{2}}[/itex] and finding the integral [itex]\int_{1}^{\infty } f(x)[/itex], but I had no luck in evaluating this by hand. Surely, I could pop it into WolframAlpha and see what it churns out, but as this is a problem in a one-variable calc textbook, I doubt that its solution demands numerical computation.
I could not find something to compare it to, though I tried taking the natural log of the product and seeing if I could compare that to anything.
[itex]\ln n e^{-n^{2}} = \ln n - n^{2} \ln e = \ln n - n^{2} > \ln n - \ln n^{2}[/itex]
So can I say therefore [itex] e^{\ln n - n^{2}} > e^{\ln n - \ln n^{2}}[/itex]?
I'm not sure if I can just exponentiate both sides of an inequality, at least not without declaring a minimum [itex]n[/itex] such that it's true, but what would the [itex]n[/itex] be greater than? If I can do that, then I have
[itex]n e^{-n^{2}} > e^{\ln \frac{n}{n^{2}}} = \frac{1}{n}[/itex], which means that the series in #13 is divergent by the Comparison Test.
For #14, I tried the a similar process, starting out by exponentiating the sequence [itex]\frac{\ln n}{n^{2}}[/itex], but I had no luck.
Any ideas?
Homework Statement
The problem says to "Determine whether the series is convergent or divergent [using only the Comparison Test or the Integral Test]."
The first, #13, is [itex]\sum_{n=1}^{\infty }\, n e^{-n^{2}}[/itex]
and the second, #14, is [itex]\sum_{n=1}^{\infty }\frac{\ln n}{n^{2}}[/itex].
Homework Equations
The Integral Test listed in Stewart says:
Suppose [itex]f[/itex] is a continuous, positive, decreasing function on [0,∞) and let [itex]a_n=f(n)[/itex]. Then the series [itex]\sum_{n=1}^{\infty }\[\itex] is convergent if and only if the improper integral [itex]\int_{1}^{\infty }f(x)dx[\itex] is convergent. In other words:
(a) If [itex] \int_{1}^{\infty }f(x)dx [/itex] is convergent, then [itex] \sum_{n=1}^{\infty }a_n [/itex] is convergent.
(b) If [itex] \int_{1}^{\infty }f(x)dx [/itex] is divergent, then [itex] \sum_{n=1}^{\infty}a_n [/itex] is divergent.
The Comparison Test listed in Stewart says
Suppose that [itex]\sum a_n[/itex] and [itex]\sum b_n[/itex] are series with positive terms.
(a) If [itex]\sum b_n[/itex] is convergent and [itex]a_n \leq b_n[/itex] for all [itex]n[/itex], then [itex]\sum a_n[/itex] is also convergent.
(b) If [itex]\sum b_n[/itex] is divergent and [itex]a_n \geq b_n[/itex] for all [itex]n[/itex], then [itex]\sum a_n[/itex] is also divergent.
The Attempt at a Solution
First, I checked to see if either sequence beings summed had a limit as [itex]n \rightarrow \infty [/itex] that was nonzero, as this would show them to be divergent, but both sequences do go to zero in the limit.
Next, I tried evaluating #13 as a function of x: [itex]f(x)=x e^{-x^{2}}[/itex] and finding the integral [itex]\int_{1}^{\infty } f(x)[/itex], but I had no luck in evaluating this by hand. Surely, I could pop it into WolframAlpha and see what it churns out, but as this is a problem in a one-variable calc textbook, I doubt that its solution demands numerical computation.
I could not find something to compare it to, though I tried taking the natural log of the product and seeing if I could compare that to anything.
[itex]\ln n e^{-n^{2}} = \ln n - n^{2} \ln e = \ln n - n^{2} > \ln n - \ln n^{2}[/itex]
So can I say therefore [itex] e^{\ln n - n^{2}} > e^{\ln n - \ln n^{2}}[/itex]?
I'm not sure if I can just exponentiate both sides of an inequality, at least not without declaring a minimum [itex]n[/itex] such that it's true, but what would the [itex]n[/itex] be greater than? If I can do that, then I have
[itex]n e^{-n^{2}} > e^{\ln \frac{n}{n^{2}}} = \frac{1}{n}[/itex], which means that the series in #13 is divergent by the Comparison Test.
For #14, I tried the a similar process, starting out by exponentiating the sequence [itex]\frac{\ln n}{n^{2}}[/itex], but I had no luck.
Any ideas?