None of these.

\sum_{n=1}^{\infty}\frac{1}{n} 

\sum_{n=1}^{\infty}\frac{1}{n^2(n+1)^2} 

\sum_{n=1}^{\infty}\frac{1}{n^2} 

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

\sum_{n=1}^{\infty}\frac{1}{\sqrt{n}} 

The determinant of A is non-zero.

|adj(A)| = 0.

adj(A) is invertible.

A has at least one non-zero entry.

With b_n=\frac{1}{n} the partial sums B_m=\sum_{n=1}^m \frac{1}{n} is bounded since m is finite.

With b_n=\sin n the partial sums |B_m|=|\sum_{n=1}^m \sin n|\leq \sum_{n=1}^m |\sin n|\leq \sum_{n=1}^m 1=m is bounded.

With a_n=\sin n, a_n is decreasing and \lim_{n\to \infty}a_n=0. 

With b_n=\sin n the partial sums

B_m=\sum_{n=1}^m \sin n=\sum_{n=1}^m \frac{\cos(n-1)-\cos(n+1)}{2\sin 1}\\=\frac{1}{2 \sin 1}(\sum_{n=0}^{m-1} \cos n-\sum_{n=2}^{m+1} \cos n)\\=\frac{1}{2 \sin 1}(\cos 0+\cos 1-\cos m-\cos (m+1))

is bounded.

With a_n=\frac{1}{n}, a_n is decreasing and \lim_{n\to \infty}a_n=0. 

With b_n=\frac{1}{n} the partial sums B_m=\sum_{n=1}^m \frac{1}{n}\leq \sum_{n=1}^m 1=m is bounded.

None of these.