Hi, in the question shown in the photo, I think I have come up with a proof for part a). However, reading through parts b) and c) I realized that the convergence of the c-series is only guaranteed if the a-series is absolutely convergent. In my proof however, I don't really use the properties of an absolutely convergent series. I am thinking I used this property in this property when I used the reverse triangle inequality without understanding. I am wondering how do I make this reliance on absolute convergence clearer in my proof, and what I should be doing for part b) and c) (I am aware of Reimann series theorem) but like I said my proof doesn't lend into that.
a) Define $A := \sum_{n=0}^{\infty} a_n$, $B := \sum_{n=0}^{\infty} b_n$.
We will now prove by induction that $\sum_{n=0}^N c_n = \sum_{k=0}^N a_k \sum_{m=0}^{N - k} b_m$. Indeed, $\sum_{n=0}^0 c_n = c_0 = a_0b_0 = \sum_{k=0}^0 a_k \sum_{m=0}^{0-k} b_m$. So we will now assume the claim is true for some arbitrary $n \in \mathbb{N}$.
Then we have $\sum_{n = 0}^{N+1} c_n = c_{N + 1} + \sum_{n = 0}^{N} c_n = \sum_{k=0}^{N+1} a_kb_{N+1-k} + \sum_{k=0}^N a_k \sum_{m=0}^{N - k} b_m = (a_0b_{N+1} + a_1b_N + \ldots + a_{N+1}b_0) + a_0(b_0 + \ldots + b_N) + a_1(b_0 + \ldots + b_{N-1}) + \ldots + a_Nb_0 = a_0(b_0 + \ldots + b_N + b_{N+1}) + a_1(b_0 + \ldots + b_N) + \ldots + a_N(b_0 + b_1) + a_{N+1}b_0 = \sum_{k=0}^{N+1} a_k \sum_{m=0}^{N+1 - k} b_m$, closing the induction.
Since the sequence of partial sums $B_N := \sum_{n=0}^{N} b_n$ converges to $B$, we know that this sequence is bounded, say by $L \in \mathbb{R}$.
Define $M := max(A, B, L)$. Take any $\epsilon > 0$ and define $\delta := \frac{\epsilon}{4M}$. Hence we can find some $N_1 \in \mathbb{N}$ such that for all $m \geq N_1$, $|\sum_{n=0}^{m} a_n - A| \leq \frac{\delta}{2}$. Therefore $|\sum_{n=0}^{m} a_n - A| = |\sum_{n=N_1 + 1}^{m} a_n - (A - \sum_{n=0}^{N_1} a_n)| \geq |\sum_{n=N_1 + 1}^{m} a_n| - |A - \sum_{n=0}^{N_1} a_n|$ by the reverse triangle inequality and since $(a_n)$. (which is proven by 'adding zero' and then applying the triangle inequality). This implies that $\frac{\delta}{2} \geq |\sum_{n=N_1 + 1}^{m} a_n - (A - \sum_{n=0}^{N_1} a_n)| \geq |\sum_{n=N_1 + 1}^{m} a_n| - |A - \sum_{n=0}^{N_1} a_n|$ so $\frac{\delta}{2} + |A - \sum_{n=0}^{N_1} a_n| \geq |\sum_{n=N_1 + 1}^{m} a_n|$ and so we can conclude that since $N_1 \geq N_1$, we have $|\sum_{n=N_1 + 1}^{m} a_n| \leq \delta$.
Now there exists some $N_2 \in \mathbb{N}$ such that for all $m \geq N_2$, $|\sum_{n=0}^{m} b_n - B| \leq \delta$. So choose $N_3 \in \mathbb{N}$ such that $N_3 - N_1 \geq N_2$. Hence, we have that for any $M \geq N_3$,$\sum_{n=0}^{M} c_n = \sum_{k=0}^M a_k \sum_{m=0}^{M - k} b_m = \sum_{k=0}^{N_1} a_k \sum_{m=0}^{M - k} b_m + \sum_{k=N_1 + 1}^M a_k \sum_{m=0}^{M - k} b_m \leq \sum_{k=0}^{N_1} a_k (B + \delta) + \sum_{k=N_1+1}^{M} a_k L \leq (A + \delta)(B + \delta) + L\delta = AB + \delta A + \delta B + \delta^2 + L\delta \leq AB + \frac{\epsilon}{4} + \frac{\epsilon}{4} + \frac{\epsilon^2}{16} + \frac{\epsilon}{4} \leq AB + \epsilon$ hence $\sum_{n=0}^{M} c_n - AB \leq \epsilon$.
Also, $\sum_{n=0}^{M} c_n = \sum_{k=0}^M a_k \sum_{m=0}^{M - k} b_m = \sum_{k=0}^{N_1} a_k \sum_{m=0}^{M - k} b_m + \sum_{k=N_1 + 1}^M a_k \sum_{m=0}^{M - k} b_m \geq \sum_{k=0}^{N_1} a_k (B - \delta) + \sum_{k=N_1+1}^{M} a_k L \geq (A - \delta)(B - \delta) - L\delta = AB - \delta A - \delta B + \delta^2 - L\delta \geq AB - \frac{\epsilon}{4} - \frac{\epsilon}{4} + \frac{\epsilon^2}{16} - \frac{\epsilon}{4} \geq AB - \epsilon$ hence $\sum_{n=0}^{M} c_n - AB \geq -\epsilon$. Hence $|\sum_{n=0}^{M} c_n - AB| \leq \epsilon$ for all $M \geq N_3$, closing the proof.