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I'm having a real hard time figuring out how to use the comparison test to check if an infinite series converges or diverges. I saw examples of solving problems like this using this test, but I still don't get it. I put two problems that I have to solve, but I have no idea where to start.

Can somebody please help me figuring this out? Thanks.

$$\sum_{m=1}^\infty \frac{1}{\sqrt{4m^2-1}}$$

$$\sum_{k=2}^\infty \frac{2^k}{3k+5}$$

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This should be asked on the main site; not on the meta site. –  Asaf Karagila Apr 7 '13 at 18:35
    
Thanks, my bad. –  user70844 Apr 7 '13 at 18:46
    
For the second series, it suffices to observe that the $n$th term does not tend to zero, so the series must diverge. If you insist on comparing, you usually can: here for instance for all sufficiently large $k$ we have $3k+5 \leq (\frac{3}{2})^k$ and thus $\frac{2^k}{3k+5} \geq \frac{2^k}{(3/2)^k} = (4/3)^k$, a divergent geometric series. –  Pete L. Clark Apr 7 '13 at 19:05

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up vote 1 down vote accepted

$1$. Note that $$\dfrac1{\sqrt{4m^2-1}} \geq \dfrac1{2m}$$ and use the fact that the harmonic series diverges to conclude that the first series diverges.

$2$. For the second one, recall that $\displaystyle \sum_{k=k_0}^{\infty} a_k$ converges if $\limsup_{k \to \infty} \dfrac{a_{k+1}}{a_k} < 1$ and diverges if $\limsup_{k \to \infty} \dfrac{a_{k+1}}{a_k} > 1$. We have $$\limsup_{k \to \infty}\dfrac{a_{k+1}}{a_k} = \limsup_{k \to \infty}\dfrac{2^{k+1}}{3(k+1)+5} \cdot \dfrac{3k+5}{2^k} = \limsup_{k \to \infty} 2 \cdot \dfrac{3k+5}{3k+8} = 2$$ Hence, the second series diverges as well.

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I got the first one, but I don't understand the second. I've never seen $\limsup_{k \to \infty}$. –  user70844 Apr 7 '13 at 18:53
    
@air_wizardo In your case, $\limsup$ is the same as $\lim$. So, do not worry too much about it, incase you are not familiar. Essentially $\limsup$ needs to be used in case the limit doesn't exist. –  user17762 Apr 7 '13 at 19:02

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