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Using the limit comparison test how do I find the test the convergence of the sum of $$\frac{1+2^{(n+1)}}{1+3^{(n+1)}}?$$

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Do you need to use the limit comparison test? The ratio test may be more suited to seeing if the sum converges. –  yunone Mar 25 '11 at 1:45
    
Or you could use the ordinary comparison test, that term being less than $\left(\frac{2}{3}\right)^n$. @Anora: Do you have any guesses as to what series to compare? –  Jonas Meyer Mar 25 '11 at 2:05

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The limit comparison test is a good substitute for the comparison test when the inequalities are difficult to establish; essentially, if you have a feel that the series in front of you is "essentially proportional" to another series whose convergence you know, then you can try to use the limit comparison test for it.

Here, for large $n$ it should be clear that $2^{n+1}+1$ is "essentially" just $2^{n+1}$; and $3^{n+1}+1$ is "essentially" the same as $3^{n+1}$. So the fraction will be "essentially", for large $n$, about the same as $\frac{2^{n+1}}{3^{n+1}}$. So this suggests using limit comparison to compare $$\sum \frac{1+2^{n+1}}{1+3^{n+1}}$$ with $$\sum\frac{2^{n+1}}{3^{n+1}} = \sum\left(\frac{2}{3}\right)^{n+1}.$$ The latter is a geometric series, so it should be straightforward to determine whether it converges or not.

So let $a_n = \frac{1+2^{n+1}}{1+3^{n+1}}$, and $b_n = \left(\frac{3}{2}\right)^{n+1}$, and compute $$\lim_{n\to\infty}\frac{a_n}{b_n}.$$ If this limit exists and is positive (greater than $0$), then both series converge or both series diverge.

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