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I would like to calculate the following limit: $$ \lim_{n\to\infty}\sum_{k=0}^{n}\alpha_k\beta_{n,k}. $$

I know that $\lim\limits_{n\to\infty}\beta_{n,k}=1$ and the series $\sum_{k\geqslant0}\alpha_k$ converges to $\ell$.

If I think about it I found the limit is $\ell$. I mean if I do something which is not math like: $$ \lim_{n\to\infty}\sum_{k=0}^{n}\alpha_k\beta_{n,k}= \left(\lim_{n\to\infty}\sum_{k=0}^{n}\alpha_k\right)\left(\lim_{n\to\infty}\beta_{n,k}\right)=\ell\cdot1=\ell.\blacksquare $$

I am sure this is wrong. Can you please help me.

P.S. As an example, one can take the following sequences:

$$ \alpha_k=\dfrac{1}{k!}. $$

and

$$ \beta_{n,k}=\dfrac{\prod\limits_{p=0}^{k-1}(n-p)}{n^k}. $$

drzbir
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1 Answers1

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What you need here is uniform convergence on the $\beta_{n,k}$'s as well as absolute convergence of the sum $\sum_{k=0}^\infty \alpha_k$. That is for each $\epsilon > 0$ there is an $N$ for which $|\beta_{n,k} - 1| < \epsilon$ for all $n > N$ and all $k$.

With this we have $$ (1-\epsilon)\sum_{k=0}^\infty |\alpha_k| \le \sum_{k=0}^\infty |\alpha_k| |\beta_{n,k}| \le (1+\epsilon)\sum_{k=0}^\infty |\alpha_k|$$

Hence $\sum_{k=0}^\infty |\alpha_k| |\beta_{n,k}|$ converges to $\sum_{k=0}^\infty |\alpha_k|$ which means $\sum_{k=0}^\infty \alpha_k \beta_{n,k}$ converges.

For your example all the terms are positive for $\alpha_k$, so the sum converges absolutely. And you should check if $\beta_{n,k}$ is converging uniformly to 1.

Joel
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