# Determining if a series converges using the comparison test

I am using the comparison test to determine if series converge. I understand how to do it when there is $1$ in the numerator:

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

$$4k^2\gt 4k^2-1$$ $$\sqrt{4k^2}\gt \sqrt{4k^2-1}$$ $$2k\gt \sqrt{4k^2-1}$$ $$\frac{1}{2k}\lt \frac{1}{\sqrt{4k^2-1}}$$

From what I understand, the first term dominates the second one, and the second one diverges, thus making the series diverge.

But I don't know where to start from on these ones:

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

$$\sum_{k=1}^\infty \frac{4^k}{3^k-1}$$

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Note that $$\frac{2^k}{3^k+5}<\frac{2^k}{3^k}$$

and $a_k=\left(\frac 2 3\right)^k$ is summable.

Can you do something similar to the other? The simple rule is that if we increase the denominator we get something smaller, and if we decrease it, something larger.

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@air_wizardo Yes, that is right – Pedro Tamaroff Apr 11 '13 at 1:16
I think your inequality sign is wrong. – user70844 Apr 11 '13 at 1:16
@air_wizardo No, it is correct. $3^k+5>3^k$ so the inequality gets reversed upon $x\mapsto x^{-1}$. For example, $5>3$ so $1/5<1/3$. – Pedro Tamaroff Apr 11 '13 at 1:17
@air_wizardo $k$ goes from $1\to \infty$ – MITjanitor Apr 11 '13 at 1:18
@PeterTamaroff: Ok, I understand how you got to this: $\frac{1}{3^k+5}<\frac{1}{3^k}$ But how do you get to what you wrote? – user70844 Apr 11 '13 at 1:22

For the last question, first take an informal look at the general term. It is big. Certainly bigger than $1$.

For the series $\sum a_k$ to converge, the terms must approach $0$. (The converse doesn't hold: in $\sum \frac{1}{k}$, the terms approach $0$ but we do not have convergence.)

In our case, the terms do not approach $0$, in fact they blow up, so the series does not converge.

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