Proving convergence of sum of two convergent series'.

Question

Suppose $$\sum_{n=1}^\infty a_n~,~\sum_{n=1}^\infty b_n$$ are convergent series' how could I use this to prove that $$\sum_{n=1}^\infty \frac{a_n+b_n}{2}$$ is also convergent using the definition for convergence of a series?

Thanks

Answer 1

Let $A$ and $B$ be the points of convergence of the two respective series. The convergence of the two series implies that given any $\varepsilon>0$, there exists an integer $N_0$ such that for any $N \ge N_0$, we have \begin{align*} \left|A-\sum_{n=1}^N a_n\right| &< \varepsilon,\\ \left|B-\sum_{n=1}^N b_n\right| &< \varepsilon.\\ \end{align*}

We claim that $\sum_{n=1}^\infty \frac{a_n+b_n}{2}$ converges to $\frac{A+B}{2}$. Indeed, for all $N \ge N_0$, we can use the triangle inequality to get $$\left|\frac{A+B}{2}-\sum_{n=1}^N \frac{a_n+b_n}{2}\right| \le \frac{1}{2}\left|A-\sum_{n=1}^N a_n\right|+ \frac{1}{2}\left|B-\sum_{n=1}^N b_n\right| < \varepsilon.$$

Answer 2

Let $S_n=\sum_{i=1}^na_i$ and $T_n=\sum_{i=1}^nb_i$ and denote $L_a=\lim S_n$ and $L_b=\lim T_n$. For a given $e>0$, there are positive integers $n_1$ and $n_2$ large enough so that $$ |S_n-L_a|<e\quad n\geq n_1,\quad |T_n-L_b|<e\quad n\geq n_2. $$ Then, with $N=\max\{n_1,n_2\}$, for all $n\geq N$, we have \begin{align*} \left|\sum_i\frac{a_i+b_i}{2}-\frac{1}{2}(L_a+L_b)\right|&=\left|\frac{1}{2}(S_n-L_a)+\frac{1}{2}(T_n-L_b)\right|\\ &\leq\frac{1}{2}|S_n-L_a|+\frac{1}{2}|T_n-L_b|<e. \end{align*} As $e$ is arbitrary, this proves $\frac{1}{2}\sum_n(a_n+b_n)\to\frac{1}{2}(L_a+L_b)$.