# Proof of Convergence by Integral Test and/or Comparison Test

1. May 30, 2012

### middleCmusic

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)

1. The problem statement, all variables and given/known data

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 $\sum_{n=1}^{\infty }\, n e^{-n^{2}}$

and the second, #14, is $\sum_{n=1}^{\infty }\frac{\ln n}{n^{2}}$.

2. Relevant equations

The Integral Test listed in Stewart says:

Suppose $f$ is a continuous, positive, decreasing function on [0,∞) and let $a_n=f(n)$. Then the series $\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$ is convergent, then $\sum_{n=1}^{\infty }a_n$ is convergent.
(b) If $\int_{1}^{\infty }f(x)dx$ is divergent, then $\sum_{n=1}^{\infty}a_n$ is divergent.

The Comparison Test listed in Stewart says

Suppose that $\sum a_n$ and $\sum b_n$ are series with positive terms.

(a) If $\sum b_n$ is convergent and $a_n \leq b_n$ for all $n$, then $\sum a_n$ is also convergent.
(b) If $\sum b_n$ is divergent and $a_n \geq b_n$ for all $n$, then $\sum a_n$ is also divergent.

3. The attempt at a solution

First, I checked to see if either sequence beings summed had a limit as $n \rightarrow \infty$ 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: $f(x)=x e^{-x^{2}}$ and finding the integral $\int_{1}^{\infty } f(x)$, 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.

$\ln n e^{-n^{2}} = \ln n - n^{2} \ln e = \ln n - n^{2} > \ln n - \ln n^{2}$

So can I say therefore $e^{\ln n - n^{2}} > e^{\ln n - \ln n^{2}}$?
I'm not sure if I can just exponentiate both sides of an inequality, at least not without declaring a minimum $n$ such that it's true, but what would the $n$ be greater than? If I can do that, then I have
$n e^{-n^{2}} > e^{\ln \frac{n}{n^{2}}} = \frac{1}{n}$, 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 $\frac{\ln n}{n^{2}}$, but I had no luck.

Any ideas?

2. May 30, 2012

Did you consider trying the substitution $u= x^2$ so that $du= 2xdx[/tex]? 3. May 30, 2012 ### micromass For 14, try to use the comparison test. Does there exist a suitable p such that [itex]ln(n)\leq n^p$ for large enough n??

4. May 30, 2012

### middleCmusic

Wow... I can't believe I didn't see that. Thanks for the help!

5. May 30, 2012

### middleCmusic

So, I tried n=0.9, and it seems to work. When graphed, n=0.9 shoots upward compared to $\ln n$ and also,
$\lim_{n \rightarrow \infty } \frac{\ln n}{n^{0.9}} = \lim_{n \rightarrow \infty} \frac{\frac{1}{n}}{0.9 n^{-0.1}} = \lim_{n \rightarrow \infty} 1.1 \frac{1}{n^{-0.9}} = 0$, so I know that as $n$ becomes infinitely large, $\ln n \leq n^{0.9}$.

I tried to prove this fact with induction, but so far can't get past the crucial step. Here's what I have so far:

Base case: $n=1$
$\ln 1=0<1^{0.9}=1$
Assume that $\ln k \leq k^{0.9}$
Now we try to show that $\ln (k+1) \leq (k+1)^{0.9}$.
If we take the 0.9 root of each side of the above inequality, we get $(\ln(k+1))^{1.1} \leq (k+1)$.
If this were a product, then I could try to separate it somehow...but as it is, I don't see how to make use of our inductive step.

If we could show this fact with induction, then we would know that $\frac{\ln n}{n^{2}} \leq \frac{n^{0.9}}{n^{2}}=\frac{1}{n^{1.1}}$ and since the last term in the (in)equality is a p-series with $p=1 \geq 1.1$, which is convergent, our series is convergent by the Comparison Test.

Thanks for all the help guys! Anyone got an idea about how to finish that induction proof?

6. May 30, 2012

### micromass

Doesn't this prove it already. I don't see why you need induction. (and I also think an induction proof would not be easy)

7. May 30, 2012

### middleCmusic

I guess you're right. It only matters that there's some $N_0$ after which the relation holds.

But I realized I could also just show that the difference of the two is always increasing... that is that $x^{0.9}- \ln x$ has a positive derivative for all n, since it already started out with $n^{0.9}$ on top.

Since $\frac{d}{dx}(x^{0.9}- \ln x)=0.9x^{-.1} - \frac{1}{x}$ and
$0.9x^{-.1} - \frac{1}{x} \geq 0$ for all $x \geq 2$, we know that the difference between them can only increase for $n \geq 2$. Thus, the inequality holds for all n, since it was proved to hold for n=1 as well.

Thanks again for the help - otherwise I might have been staring at this one for a while. :P

8. May 31, 2012

### algebrat

Integration by parts for ln(x)/x^2.

Let u=ln(x), dv=dx/x^2.

cannot delete accidental repeat?? how google of you

Last edited: May 31, 2012
9. May 31, 2012

### algebrat

Integration by parts for ln(x)/x^2.

Let u=ln(x), dv=dx/x^2.

By the way, I can look at these and know they're convergent in about five seconds, in my head. The exponential goes faster than any polynomial. Natural log goes slower than x. And anything that goes faster than 1/n converges. Be careful with this rule. At least you can guess faster. And it might have given you the idea micromass gave you.

10. May 31, 2012

### middleCmusic

Thanks for the tip! Your method is way simpler than what I was trying to come up with and I'm sure it will be useful.

Just to be sure I know what you're saying, for these two examples, we have

#13 Since $e^{n^{2}} > n^{2}$, we must have that $\frac{n}{e^{n^{2}}} < \frac{n}{n^2} = \frac{1}{n}$ and so it's basically equivalent to a convergent p-series with $p>1$

#14 And for this one, $\ln n < n$, so $\frac{\ln n}{n^{2}} < \frac{n}{n^{2}} = \frac{1}{n}$, so this one also is equivalent to a convergent p-series with $p>1$.

Thanks again. :)

11. May 31, 2012

### algebrat

Mmm, careful, you are showing it is less than a divergent series. Just to clarify, this is very informal reasoning, and proofs should be formal lest they prove something false.

For instance, 1/(2n)<1/n, or 1/(n+1)<1/n. The reasoning applied carelessly would not work here. It is a trick for developing an initial conjecture with better speed. With experience, you can get good at using this "rule".

Try it on (n^2+2n)/(n^3*ln(n)+60n). Does its series converges?

Last edited: May 31, 2012