Special Comparison Test For Infinite Series

[Ascendant0]

Homework Statement

Use the special comparison test to find whether the series converges or diverges:
$\sum_{n=5}^{\infty}1/({2^n-n^2})$

Relevant Equations

If $\sum_{n=1}^{\infty}b_n$ is a convergent series of positive terms and $a_n \geq 0$ and $a_n/b_n$ tends to a finite limit, then $\sum_{n=1}^{\infty}a_n$ converges

Obviously, you can tell from the fraction that it converges. My problem is their explanation of this process in the book is extremely convoluted, so I'm not too sure what to do with this?

From what I gather from their example in the book, I'd want to first create $b_n$ out of the "important values of this equation as $n \rightarrow \infty$, which since as it approaches $\infty$, the important piece is $1/(2^n)$. Even more obvious now that it converges...

However, now I'm supposed to set $a_n/b_n$, which ends me up with $2^n/(2^n-n^2)$. Then, they simplified their problem by dividing both the numerator and the denominator by the highest term. With theirs though, it was easy, as they had $n^2 \sqrt{2n^2}/(4n^3)$, so it simply reduces down to $\sqrt{2}/4$, and so it's finite and hence converges.

Conversely, I'm not even sure what to do with the equation I end up with? With an $n$ in the exponent, I feel I'd want to take the $ln$ of the top and bottom of the fraction to get the $n$, but then I end up with $ln(2^n-n^2)$ in the denominator, and I don't know any ln rules that would let me split/simplify that.

I mean could I say that the fraction I end up with from $a_n/b_n$ "tends to a finite limit," but they further simplified their equation before completing the problem.

Also, I know the $-n^2$ causes the denominator to always be a lower value, that doesn't matter because of how insignificant it is compared to the other two terms, right? Because it only has to "tend" to a finite limit, not actually "become" a finite limit? Is that all I need to do here, because I'm really not 100% sure this is exactly how this is done?

 

[anuttarasammyak]

I expect : There exists M such that for n>M
2^n-n^2 > 2^{n-1}
\frac{1}{2^n-n^2} < \frac{1}{2^{n-1}}
\sum_{n=M}^\infty \frac{1}{2^{n}} < \sum_{n=M}^\infty \frac{1}{2^n-n^2} < \sum_{n=M}^\infty \frac{1}{2^{n-1}}
RHS converges to a finite value.

 

[Ascendant0]

I expect : There exists M such that for n>M
2^n-n^2 > 2^{n-1}
\frac{1}{2^n-n^2} < \frac{1}{2^{n-1}}
\sum_{n=M}^\infty \frac{1}{2^{n}} < \sum_{n=M}^\infty \frac{1}{2^n-n^2} < \sum_{n=M}^\infty \frac{1}{2^{n-1}}
RHS converges to a finite value.

Thank you, and we have done that type of thing previously, the regular "Comparison Test." These problems in this section ask specifically for the "Special Comparison Test," where you're supposed to break $a_n$ down to it's important parts as $n \rightarrow \infty$, and then divide $a_n$ by the new equation made from breaking it down, like the example problem they gave in the book shown below.

I couldn't include all of it, as it's on two separate pages with a description in between, but the values to the left of the division sign is the $a_n$ the example is testing, and the $1/n^2$ on the right side of it is what they broke the important values of $a_n$ down to as it approaches infinity. Essentially, just the leftover $n$ values of the highest terms after you reduce them as much as possible.

The thing is though, I see all their other terms tend to "0" as $n \rightarrow \infty$, whereas in my equation I set up, the "$-n^2$" does not. I'm not sure if that is an issue for this kind of testing proof and needs to be reduced further somehow (or handled differently altogether), or if since the dominant value will be the $2^n$ in my problem, if that becomes irrelevant and my answer is sufficient?

 

[pasmith]

Homework Statement: Use the special comparison test to find whether the series converges or diverges:
$\sum_{n=5}^{\infty}1/({2^n-n^2})$
Relevant Equations: If $\sum_{n=1}^{\infty}b_n$ is a convergent series of positive terms and $a_n \geq 0$ and $a_n/b_n$ tends to a finite limit, then $\sum_{n=1}^{\infty}a_n$ converges

However, now I'm supposed to set $a_n/b_n$, which ends me up with $2^n/(2^n-n^2)$. Then, they simplified their problem by dividing both the numerator and the denominator by the highest term. With theirs though, it was easy, as they had $n^2 \sqrt{2n^2}/(4n^3)$, so it simply reduces down to $\sqrt{2}/4$, and so it's finite and hence converges.

Conversely, I'm not even sure what to do with the equation I end up with? With an $n$ in the exponent, I feel I'd want to take the $ln$ of the top and bottom of the fraction to get the $n$, but then I end up with $ln(2^n-n^2)$ in the denominator, and I don't know any ln rules that would let me split/simplify that.

I mean could I say that the fraction I end up with from $a_n/b_n$ "tends to a finite limit," but they further simplified their equation before completing the problem.

Cancel a common factor of 2^n: \begin{split}<br /> \frac{a_n}{b_n} &amp;= \frac{2^n}{2^n - n^2} \\<br /> &amp;= \frac{1}{1 - \frac{n^2}{2^n}} \end{split} Now if n^2/2^n tends to a limit L \neq 1 then a_n/b_n tends to (1 - L)^{-1}. And there are many ways of showing that \lim_{n \to \infty} n^2/2^n = 0.

Also, I know the $-n^2$ causes the denominator to always be a lower value, that doesn't matter because of how insignificant it is compared to the other two terms, right? Because it only has to "tend" to a finite limit, not actually "become" a finite limit?

Nothing "becomes" anything. \lim_{n \to \infty} \frac{2^n}{2^n - n^2} is finite; \frac{2^n}{2^n - n^2} tends to a finite limit as n tends to infinity.

 

[Ascendant0]

Cancel a common factor of 2^n: \begin{split}<br /> \frac{a_n}{b_n} &amp;= \frac{2^n}{2^n - n^2} \\<br /> &amp;= \frac{1}{1 - \frac{n^2}{2^n}} \end{split} Now if n^2/2^n tends to a limit L \neq 1 then a_n/b_n tends to (1 - L)^{-1}. And there are many ways of showing that \lim_{n \to \infty} n^2/2^n = 0.

Nothing "becomes" anything. \lim_{n \to \infty} \frac{2^n}{2^n - n^2} is finite; \frac{2^n}{2^n - n^2} tends to a finite limit as n tends to infinity.

Thank you for the clarification. I get it now.