Prove or give a counterexample.
If \(\sum a_n\) and \(\sum b_n\) converge, then \(\sum(a_n+b_n)\) converges.
If \(\sum a_n\) and \(\sum b_n\) converge, then \(\sum a_nb_n\) converges.
Solution
1. Statement (a) is true.
Let \[ A_N\coloneqq\sum_{n=1}^N a_n, \qquad B_N\coloneqq\sum_{n=1}^N b_n. \] By hypothesis there are real numbers \(A,B\) such that \[ A_N\to A, \qquad B_N\to B. \] The \(N\)th partial sum of \(\sum(a_n+b_n)\) is \[ \sum_{n=1}^N(a_n+b_n) = A_N+B_N. \] Therefore \[ A_N+B_N\longrightarrow A+B, \] so \(\sum(a_n+b_n)\) converges.
2. For\[ a_n=b_n=\frac{(-1)^n}{\sqrt n} \qquad(n\ge1), \]both series\[ \sum_{n=1}^{\infty}a_n \qquad\text{and}\qquad \sum_{n=1}^{\infty}b_n \]converge.
The positive sequence \[ \frac1{\sqrt n} \] is decreasing and tends to \(0\). Hence \[ \sum_{n=1}^{\infty}\frac{(-1)^n}{\sqrt n} \] converges by the alternating-series test. Since \(a_n=b_n\), both stated series converge.
3. For the sequences in step 2, the termwise-product series\(\sum_n a_nb_n\) diverges.
For every \(n\), \[ a_nb_n = \frac{(-1)^{2n}}{n} = \frac1n. \] Thus \[ \sum_{n=1}^{\infty}a_nb_n = \sum_{n=1}^{\infty}\frac1n, \] which is the divergent harmonic series. Therefore statement (b) is false.
4. Q.E.D.
Step 1 proves (a), while steps <1>2–<1>3 give a counterexample to (b).