Series comparisons and logarithms? Prove the convergence of
$$
\sum_{n = 1}^{\infty}
{\sqrt{\, 2n - 1\,}\,\ln\left(4n + 1\right) \over n\left(n+1\right)}
$$
I've been struggling for hours on this. By the textbook we have the limit test, comparison test, asymptotic test and integral test. When I use the asymptotic test my quotient ends up as $0$ or $\infty$, and I cannot prove anything. Since I am self studying, I have no one to turn to. The rest of the questions were easy, but I cant figure this out. Please help!
 A: The two problems yield to basically the same technique. We look at the first one, which is somewhat harder.
First let us argue informally. Note that $\sqrt{2n-1}$ in the long run behaves like $\sqrt{2}\sqrt{n}$. Since we won't worry about constants, let us say it behaves like $\sqrt{n}$. 
The bottom behaves like $n^2$, so $\frac{\sqrt{2n-1}}{n(n+1)}$ behaves like $\frac{1}{n^{3/2}}$.  
The term $\ln(4n+1)$, in the long run, grows very slowly, more slowly than any power of $n$. So in the long run, in particular, it grows more slowly than $n^{1/4}$. Thus, in the long run, our original expression goes to $0$ faster than $\frac{1}{n^{5/4}}$. 
But $\sum_1^\infty \frac{1}{n^{5/4}}$ converges. It follows that our original series converges. 
In essence, in $\frac{1}{n^{3/2}}$, the exponent $3/2$ is plenty big enough to ensure convergence. So we grab an $n^{1/4}$ from it to "kill" the $\ln(4n+1)$ term. That leaves us with $\frac{1}{n^{5/4}}$, which still goes to $0$ fast enough. 
Now it is time to turn the above into a more formal argument, say by using the Limit Comparison Test. Let $a_n$ be the $n$-th term of our series. We show that
$$\lim_{n\to\infty} \frac{a_n}{\frac{1}{n^{5/4}}}=0.$$
Since $n^{5/4}=n^{3/2}/n^{1/4}$, we want to show that 
$$\lim_{n\to\infty} \frac{\sqrt{2n-1}\,n^{3/2}}{n(n+1)}\frac{\ln(4n+1)}{n^{1/4}}=0.$$
We can use L'Hospital's Rule to show that $\lim_{n\to\infty}\frac{\ln(4n+1)}{n^{1/4}}=0$.
Standard techniques show that 
$$\lim_{n\to\infty}\frac{\sqrt{2n-1}\,n^{3/2}}{n(n+1)}=\sqrt{2}.$$
This completes the Limit Comparison argument. 
A: I would suggest you start with some rough hand-wavy approximations and then make it rigorous. For example, when $n$ is large,
$$\frac {\sqrt{2n-1}}{n(n+1)}\approx \frac {\sqrt {2n}}{n^2}=\sqrt 2 n^{-3/2}$$
and $\ln (4n+1) \sim \ln n$.
Let's try integrating
$$\int x^{-3/2} \ln x \,dx.$$
Let $u = \ln x$, $du = \frac 1 x dx$, $dv = x^{-3/2}$, $v=-2 x^{-1/2}$.
This gives an integral of
\begin{align*}
-2x^{-1/2}\ln x - \int -2x^{-3/2}\,dx
&=-2x^{-1/2}\ln x-4x^{-1/2} +C\\
&=-2\frac{\ln x}{\sqrt x}-4\frac{1}{\sqrt x}+C.
\end{align*}
So the integral from, say, $1$ to $+\infty$ of this thing is quite clearly finite.
A: $\ln n$ grows much, much, much more slowly than any positive power of $n$.  So for $n$ large enough,
$$ \frac{\sqrt{2n-1}\ln(4n+2)}{n(n+1)} < \frac{\sqrt{2n}*n^{1/10}}{n^2}<\frac{2}{n^{14/10}}.$$
Use the Comparison Test (then the Integral Test).  You may have to justify the first inequality, depending on who's asking.
