3

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:

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

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

2 Answers2

1

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.

Pedro
  • 122,002
  • @air_wizardo Yes, that is right – Pedro Apr 11 '13 at 01:16
  • I think your inequality sign is wrong. –  Apr 11 '13 at 01:16
  • @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 Apr 11 '13 at 01:17
  • @air_wizardo $k$ goes from $1\to \infty$ – MITjanitor Apr 11 '13 at 01:18
  • @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? –  Apr 11 '13 at 01:22
  • @MITjanitor: That's not what my problem says. –  Apr 11 '13 at 01:23
  • @air_wizardo If $a<b$ and $c>0$, $ac<bc$. In this case $c=2^k$ which is always positive. For example, $2<4$ and $2>0$, so $2\cdot 2<2\cdot 4$, i.e. $4<8$. – Pedro Apr 11 '13 at 01:23
  • @PeterTamaroff: Not sure if I understand. That's ok, thanks for the help. –  Apr 11 '13 at 01:25
  • @air_wizardo What is it you don't understand? Don't you know the rules/axioms of order in $\Bbb R$? We have that $$\frac{1}{a}<\frac{1}{b}$$ Since $c>0$ get that $$\frac ca<\frac cb$$ – Pedro Apr 11 '13 at 01:26
  • @PeterTamaroff: Don't worry about it. It's late and my brain doesn't work. –  Apr 11 '13 at 01:27
  • @air_wizardo That's what you wrote above... – MITjanitor Apr 11 '13 at 01:45
1

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.

André Nicolas
  • 507,029