Prove convergence of a series little with Direct comparison test I have the following series
$$\sum_{k=1}^{\infty} \log\left(1+\frac{1}{k^2}\right)$$
This series should converge but when I apply the Direct comparison test it diverges
$$\left|\sum_{k=1}^{\infty} \log(1+\frac{1}{k^2})\right| \le \sum_{k=1}^{\infty} \log\left|1+\frac{1}{k^2}\right| \le \sum_{k=1}^{\infty} 1+\frac{1}{k^2} = \sum_{k=1}^{\infty} 1 + \sum_{k=1}^{\infty}\frac{1}{k^2}$$ 
so we know that $\sum_{k=1}^{\infty} \left(1+\frac{1}{k^2}\right) = \sum_{k=1}^{\infty} 1 + \sum_{k=1}^{\infty}\frac{1}{k^2}$ diverges because $\sum 1$ diverges, so the series should diverges.
what am I doing wrong?
Should it be in this way :
$$\left|\sum_{k=1}^{\infty} \log(1+\frac{1}{k^2})\right| \le \sum_{k=1}^{\infty} \log\left|1+\frac{1}{k^2}\right| \le \sum_{k=1}^{\infty} \frac{1}{k^2}$$
so it will converges because $\sum \frac{1}{k^2}$ converges ? if yes, why do we ignore $\sum 1$
 A: Remember the comparison test:

Given two real sequences $(a_n)$ and $(b_n)$,
  
  
*
  
*If $0\le a_n \le b_n$ and $\sum_{n=1}^{\infty} b_n$ converges, then $\sum_{n=1}^{\infty}a_n$ converges.
  
*If $0\le b_n \le a_n$ and $\sum_{n=1}^{\infty} b_n$ diverges, then $\sum_{n=1}^{\infty}a_n$ diverges.
  

To show that $\sum \log\left(1+\frac{1}{k^2}\right)$ diverges, then you have to find another sequence $(c_n)$ such that $0\le c_n \le a_n$ and show $\sum c_n$ diverges. However, you did a wrong way, finding $\log\left(1+\frac{1}{k^2}\right)\le 1+\frac{1}{k^2}$. $\sum (1+(1/k^2))$ diverges, but it gives no reason to apply comparison test.
Actually, $\sum \log\left(1+\frac{1}{k^2}\right)$ converges. It is known that following inequality holds for $x>0$:
$$
0\le\log(x+1)\le x
$$
Substitute $x=\frac{1}{k^2}$, then we get
$$
0\le\log\left(1+\frac{1}{k^2}\right)\le \frac{1}{k^2}
$$
and $\sum \frac{1}{k^2}$ converges. By comparison test, $\sum \log\left(1+\frac{1}{k^2}\right)$ converges.
A: Since $\log(1+x)\le x$ for all $x\gt-1$, we have
$$
\begin{align}
\sum_{k=1}^\infty\log\left(1+\frac1{k^2}\right)
&\le\sum_{k=1}^\infty\frac1{k^2}\\
&=\frac{\pi^2}6\tag{1}
\end{align}
$$
so the sum converges by comparison.

Since $\log(1+x)\ge\frac{x}{1+x}$ for all $x\gt-1$, we have
$$
\begin{align}
\sum_{k=1}^\infty\log\left(1+\frac1{k^2}\right)
&\ge\sum_{k=1}^\infty\frac1{k^2+1}\\
&=\frac\pi2\coth(\pi)-\frac12\tag{2}
\end{align}
$$
In fact,
$$
\begin{align}
\prod_{k=1}^\infty\frac{k^2+1}{k^2}
&=\lim_{n\to\infty}\prod_{k=1}^n\frac{k+i}{k}\frac{k-i}{k}\\
&=\lim_{n\to\infty}\frac{\Gamma(n+1+i)}{\Gamma(1+i)}\frac{\Gamma(n+1-i)}{\Gamma(1-i)}\frac1{\Gamma(n+1)^2}\\
&=\frac1{i\,\Gamma(i)\,\Gamma(1-i)}\lim_{n\to\infty}\frac{\Gamma(n+1+i)\Gamma(n+1-i)}{\Gamma(n+1)^2}\\
&=\frac{\sin(\pi i)}{\pi i}\cdot1\\[3pt]
&=\frac{\sinh(\pi)}\pi\tag{3}
\end{align}
$$
Therefore,
$$
\sum_{k=1}^\infty\log\left(1+\frac1{k^2}\right)
=\log\left(\frac{\sinh(\pi)}\pi\right)\tag{4}
$$
As shown in $(1)$ and $(2)$,
$$
\frac\pi2\coth(\pi)-\frac12
\lt\log\left(\frac{\sinh(\pi)}\pi\right)
\lt\frac{\pi^2}6\tag{5}
$$
