# Selection of $b_n$ in Limit Comparison Test for checking convergence of a series

I wonder what the correct criterion for selection of $b_n$ in Limit Comparison Test for checking convergence of a series. Any hint to online material will be highly appreciated. The selection of $b_n$ in first series is easy but in other three is tricky. Is there any universal criterion for selection of $b_n$? Thanks in advance for your help.

$\sum_{1}^{\infty}\frac{\sqrt{n}}{1+n}$ then $b_{n}=\frac{\sqrt{n}}{n}=\frac{1}{\sqrt{n}}$

$\sum_{1}^{\infty}\frac{\ln n}{n}$ then $b_{n}=\frac{1}{n}$

$\sum_{1}^{\infty}\sin(\frac{\pi}{n})$ then $b_{n}=\frac{\pi}{n}$

$\sum_{1}^{\infty}\frac{\ln\left(n+1\right)}{n^{2}}$ then $b_{n}=\frac{1}{n^{\frac{3}{2}}}$

-

## 1 Answer

There's no universal condition, no. The broadest hint I can give is that geometric series are easy (they converge under nice conditions), and the harmonic series is too (it diverges nicely). So your goal should be "compare to something that gets you a little closer to one of those." For instance, for $3n / (1 + n^2)$, you can look and say "as $n$ gets large, that kinds looks like $1/n$. Let's try that."

There are usually two flavors of "let's try that."

One is to say, "Hey, each term is a little less than $3/n$, so we'll use $3/n$ for comparison. That's a good idea if the thing you're comparing to CONVERGES, but useless in this case, since the harmonic series diverges.

The other is to say "Hey, if it's close to $3/n$, it must be bigger than $2/n$ eventually." So you optimistically write down

$\frac{3n}{1+n^2} \ge 2/n$

and do some algebra to get $$3n^2 \ge 1 + n^2 \\ 2n^2 \ge 1.$$

and conclude that for $n$ at least one, your comparison is good. Sometimes you do this and find that your comparison is valid for $n >10$, say. When that happpens, you say that $$\sum_{n=1}^\infty a_n = \sum_{n=1}^{10} a_n + \sum_{n=11}^\infty a_n$$ and then estimate the last sum (the second one being finite, hence of no concern) with your comparison.

Hope this vague set of notions is of some help to you.

Let me add two more ideas. One is that for things like $\sin(\pi/n)$, the argument to the function often heads for some obvious limit (in this case, $\pi/n \to 0$, but in the case $\ln \frac{n^2}{n^2 + 1}$, the argument heads towards $1$). You can often take the outside function ($\sin$ in the first case, $\ln$ in the second) and approximate it with a Taylor series, or, more simply, with a linear approximation. For $\sin$, this is $$\sin x \approx x \text{, for x \approx 0}.$$ For $\ln$, it's $$\ln x \approx x -1 \text{, for x \approx 1}$$

You can then "simplify" things by saying that $\sin(\pi/n) \approx \pi/n$, or $$\ln \frac{n^2}{n^2 + 1} \approx \frac{n^2}{n^2 + 1} - 1 = \frac{-1}{n^2 + 1},$$ and use whatever you get as your comparison series $\{b_n\}$.

The second idea is that certain functions can be replaced by other simpler things. For instance, iff you have $$a_n = \frac{1}{n^2 + \sin n},$$ it looks terrible. But once you realize that $\sin(n)$ is between $-1$ and $1$, and for $n > 10$, say, this doesn't matter a bit, you can say, "Let's just look at the series for $n > 10$, and write $$\frac{1}{n^2 + \sin n} \le \frac{1}{\frac{n^2}{2}}$$ because $\sin n$ is certainly smaller than $\frac{n^2}{2}$." Similarly, it's not a bad idea to pretend in your head that $\ln n$ is pretty nearly $1$, at least to get a sense of where things are going. Alas, for the most sensitive stuff, that loss of a $\ln n$ factor will be exactly the breaking point, but often it's a good way to get a general feel for what's happening in more clearly convergent or divergent things that just happen to be algebraically complicated.

-