The limit comparison test I have a serious problem with this problem.
$$ \sum^{\infty}_n \frac{\frac{1}{n} + n}{\sqrt{\log{n} + n^5}}$$
I know that the series of this converges. However I have to use the limit comparison test. I am so confused that i don't even know where to start. I know $A_n= \dfrac{1/n+n}{\sqrt{\log(n)+n^5}}$ but $B_n= \dfrac1{n^4}$?
Can anyone help me step by step?
 A: For all $n>1$, we have $$\dfrac{1/n+n}{\sqrt{n^5 + \log(n)}} < \dfrac{n+1/n}{\sqrt{n^5}} = \dfrac1{n^{3/2}} + \dfrac1{n^{7/2}}$$
Hence,
$$\sum_{n=1}^{\infty}\dfrac{1/n+n}{\sqrt{n^5 + \log(n)}} < \sum_{n=1}^{\infty} \dfrac1{n^{3/2}} + \sum_{n=1}^{\infty} \dfrac1{n^{7/2}} = \zeta(3/2) + \zeta(7/2) < \infty$$
A: Before you use the limit comparison test, one needs to first decide what to compare your series to. This requires a little bit of intuition, but there are several key principles that can guide you in making your choice. One should remember the following hierarchy:

Factorial > exponential > positive power > logarithm.
  Higher power > lower power.

The key here is that all we really care about when we figure out whether the series converges is what happens when $n$ is extremely large. So when that happens, all we want to look at is the things that are the most important for large $n$. In our case, we have series terms
$$ A_n = \frac{\frac{1}{n} + n}{\sqrt{\log{n} + n^5}} $$
In the numerator, the dominant term is $n$, and in the denominator, the dominant term inside the square root is $n^5$, so for large $n$, all that matters is $n/\sqrt{n^5} = n^{-3/2}$. Thus the series you should use to compare with is
$$\sum \frac{1}{n^{3/2}}$$
A: Define
$$a_n:=\frac{\frac{1}{n}+n}{\sqrt{\log n+n^5}}\Longrightarrow \frac{a_n}{\frac{1}{n^{3/2}}}=\frac{\sqrt n+n^{5/2}}{\sqrt{\log n+n^5}}\xrightarrow[n\to\infty]{}1$$
and thus our series converges (why?)
