Does $\lim_{n\to\infty} b_{n,k}=1$ imply $\lim_{n\to\infty}\sum_{k=1}^n a_kb_{n,k}=\sum_{n=1}^\infty a_n$?

Question

From the reasoning in the answer to this question, I think that if a series $\sum a_n$ of positive terms converges and for each $1\le k\le n$, we have a number $0<b_{n,k}\le1$, and for each $k$ it is $\lim_{n\to\infty} b_{n,k}=1$, then $$\sum_{n=1}^\infty a_n=\lim_{n\to\infty}\sum_{k=1}^n a_kb_{n,k}$$ Is that correct? If it is, can be the conditions about $b_{n,k}$ be relaxed in any way?

Answer 1

It suffices that $(a_n)$ is absolutely convergent, $\lim_{n\to\infty} b_{n,k}=1$ for all $k$, and that $(b_{n,k})$ is bounded.

Then $|a_n| \le K$, $|b_{n,k}| \le M$, and for $1 < N < n$ is $$ \left|\sum_{k=1}^n a_k-\sum_{k=1}^n a_kb_{n,k} \right| \le \sum_{k=1}^n |a_k| |1-b_{n,k}| \le K \sum_{k=1}^N |1-b_{n,k}| + (1+M) \sum_{k=N+1}^n |a_k| \, . $$ Now, given $\epsilon > 0$, one can first choose $N > 0$ such that $\sum_{k=N+1}^n |a_k| < \frac{\epsilon}{2(1+M)}$, and then choose $n_0 > N$ such that $|1-b_{n,k}| < \frac{\epsilon}{2KN}$ for $1 \le k \le N$ and $n \ge n_0$. It follows that $$ \left|\sum_{k=1}^n a_k-\sum_{k=1}^n a_kb_{n,k} \right| < \epsilon $$ for $n\ge n_0$, so that the limits are equal.

This can also be shown with the help of the dominated convergence theorem for integrals with respect to the counting measure on the positive integers: Let $f, f_n: \Bbb N \to \Bbb R$ be the functions defined by $$ \begin{align} f(k) &= a_k \, ,\\ f_n(k) &= \begin{cases} a_k b_{n,k} &\quad \text{ if } k \le n \, ,\\ 0 &\quad\text{ if } k > n \, , \end{cases} \end{align} $$ and $\mu$ be the counting measure on $\Bbb N$. Then $|f_n(k)| \le M |f(k)|$ and $f$ is integrable, so that the DCT can be applied: $$ \sum_{k=1}^n a_kb_{n,k} = \int_{\Bbb N} f_n(k) d\mu \to \int_{\Bbb N} f(k) d\mu = \sum_{k=1}^\infty a_k $$ for $n \to \infty$.

If all $a_n$ and $b_{n,k}$ are non-negative and $b_{n,k}$ increases to $1$ for each fixed $k$, then monotone convergence theorem (Beppo Levi's lemma) can be applied. It follows that under these conditions, the identity holds even if $\sum_{n=1}^\infty a_n = \infty$.