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I am taking a course in analysis and would just like to clarify my understanding with this motivating example?

The series is:

$$ \sum_{n=0}^\infty \frac{2^n - 1}{3^n + 1} $$

My textbook defines the limit comparison test as: if sequences $0<a_k$ and $0<b_k$ and limSup$\frac{a_n}{b_n} < \infty$

Then if $\sum_{n=0}^\infty b_k$ converges $\implies$ $\sum_{n=0}^\infty a_k$ converges and if $\sum_{n=0}^\infty a_k$ diverges then $\sum_{n=0}^\infty b_k$ diverges.

My question is this: when evaluating limSup$\frac{a_n}{b_n}$, am I correct that I should first determine the Limit Superior of $\frac{a_n}{b_n}$ and then take the limit as n goes to infinity? Because then, for this question, comparing with $\sum_{n=0}^\infty \frac{2^n}{3^n}$ I get:

limSup$\frac{a_n}{b_n} $ = $\lim_{n \to \infty} (Sup_{k \ge n} \frac{3^n}{2^n}\cdot\frac{2^n - 1}{3^n + 1} ) = \frac{8}{7}$, when n = 2

(The supremum is $\frac{8}{7}$ by evaluating some terms (I realise this will not always work) and is independent of $n$).

But from some examples I have seen, I seem to be able to just manipulate above to get:

$\lim_{n \to \infty} (Sup_{k \ge n} \frac{1-\frac{1}{2^n}}{1}\cdot\frac{1}{1-\frac{1}{3^n}} ) = 1$ as $n$ tends to infinity.

Obviously, these two operations are not the same. Which one is correct (if any!)? Could someone, help me to clarify my confusion?

Thanks so much!

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    $\begingroup$ Too fancy. The $n$-th term is less than $2^n/3^n$, so straight Comparison. $\endgroup$ – André Nicolas Mar 30 '14 at 21:26
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You want to compare this series to $\displaystyle\sum_{n=1}^{\infty} \frac{2^n}{3^n}$, and here the regular Comparison Test will work, since $\;\;\;\;\displaystyle\frac{2^n-1}{3^n+1}<\frac{2^n}{3^n}=\bigg(\frac{2}{3}\bigg)^n$.

If you want to use the Limit Comparison Test, though, then you can use your second computation, since $\displaystyle\lim_{n\to\infty}\frac{2^n-1}{3^n+1}\div\frac{2^n}{3^n}=\lim_{n\to\infty}\frac{1-1/2^n}{1+1/3^n}=1$ and therefore $\displaystyle\limsup_{n\to\infty}\frac{2^n-1}{3^n+1}\div\frac{2^n}{3^n}=1$.

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$\frac{2^n - 1}{3^n + 1} \leq \frac{2^n}{3^n + 1} \leq \frac{2^n}{3^n}$

bigger one converges so is smaller.

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Honestly I'm confused with your definitions.

First, you are confronting with the wrong series ($\frac{3^n}{2^n}$ does not converge, so it does not tell you much)

Plus, in this case another test is better; it states that if $f(x) \sim g(x)$, then $\sum f(x) $ converges if and only if $\sum g(x)$ does.

In your case, the sum is $\sim \sum \frac{2^n}{3^n} $ that clearly converges (root test, if you want :-) )

Now, regarding the supremum limit, I dont' understand your definitions; if a functions admits limit (and $\frac{3^n}{2^n}$ does) then the supremum limit (and the inferior limit) are all equal to the "usual" limit. (in this case $\infty$)

What is exactly your definitions of supremum limit?

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  • $\begingroup$ Sorry, the comparison series was a typo. I've clarified my confusion now. Many thanks! $\endgroup$ – JackReacher Mar 31 '14 at 5:12

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