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Librodecalculo4000pdf

A:

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31 May, 2016. Pdf Kit: Similar Subjects. Advanced Other Comments. Calc Dif Calc Poly Calc Potencial Calc Rectangular Vesica. Calculo Calculo Manual Calculo Endcilixa Special Geometric.Q:

Is an arbitrary series of nested summations that converges absolutely convergent?

I have never before seen the notation for the following sum.
$$\sum_{i=0}^\infty {i+1 \choose i}x^i\sum_{i=0}^\infty i!x^i$$
The second summation does not seem to converge as the outer sums diverge. So is it possible that the sum converges?

A:

The last term in the partial sum is
$$x^i i! \leq x^i i^2 \leq x^i e^{i^2}=\frac{1}{1-x}$$
and it can be shown (using comparison test or divergence test) that the series $S=\sum_{i=0}^\infty \frac{x^i}{1-x}$ is convergent.

A:

Note that the nested sum is in fact the exponential generating function of the Catalan numbers. By the Lagrange inversion theorem,
\begin{align*}
\sum_{i=0}^\infty {i+1 \choose i} x^i i! &= \sum_{i=0}^\infty \frac{x^i}{i!} \sum_{j=0}^\infty {j+i \choose j} x^j = \sum_{n=0}^\infty C_n x^n = e^{x+\frac{x^2}2+\frac{x^3}3+\ldots} = \sum_{i=0}^\infty \frac{x^i}{(1-x)^i},
\end{align*}
which should be compared with the series
$$
\sum_{i=0}^\infty \frac{x^i}{(1-x)^i} = \sum_{i=0}^\infty (1-x)^i = \frac1{1-(1-x)} = \
3da54e8ca3

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