# “Weighted” Summation

Since already two utents have misunderstood the question, I'm pointing it out: I've already proved the first problem, I'm using it just as an example and as motivation to the question, which is the one quoted

Trying to prove that

$$$\lim_{n\to \infty}\left(1+\frac{x}{n}\right)^n=\sum\limits_{k=0}^{\infty}\frac{x^k}{k!}$$$

I founded this expression:

$$$\lim_{n\to \infty}\sum\limits_{k=1}^n\frac{x^k}{k!}\frac{n(n-1)\dots(n-k)}{n^k}$$$

Prove the equivalence is possible since we can in this case use the dominated convergence, since both the series and the weight are "nice", but these led my to wondering:

Under wich condition, given that \begin{align}\forall k \lim_{n\to \infty}b_{n,k}=1\\ \sum\limits_{k<n}a_k b_{n,k} \to \sum\limits_k a_k\end{align}

I tried to use the dominated convergence theorem in general, as I did in proving the first problem, with a discrete measure on $\mathbb{N}$, but I always ended up with hypothesis that seemed to me stronger than the statement. I managed to show that $\sum\limits_{k=2}^n\frac{k^{\frac{2}{n}}}{n\ln^2(k)}$ diverges, while $\sum\limits_{k=2}^\infty\frac{1}{n\ln^2(n)}$ converges absolutely, so I think there must be some kind of hypothesis on the rate of growth of $b_{n,k}$, but I do not know how to make these ideas precise

Obviously, this is strictly related to summation's method (as the Cesàro one), but I would like to prove it relazing the condition on the average of the weights that I've always found in this topic. More, it would be a useful generalization, for example of Convergence of a modified series. Any bibliography is really welcome too

• You could prove that both converge to $e^x$, but are you looking for a proof that does not use $e^x$? – dezdichado Feb 2 '18 at 19:43
• @dezdichado The convergence of the first sum I came up with it's not the matter here, the question is the one quoted – Gabriele Cassese Feb 2 '18 at 19:45
• Not for the series but rather for the 'averaged limit', but there is a characterization theorem for the latter, such as the Silverman-Toeplitz theorem – Sangchul Lee Mar 11 '18 at 5:15
• Dominated convergence is the way to go. – Antonio Vargas Mar 11 '18 at 5:17
• @AntonioVargas Isn't there a weaker method? Assuming dominated convergence seems to much to me – Gabriele Cassese Mar 11 '18 at 6:59