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Some hints. The sum indicized on $\mathbb{N}^{< \omega}$ would like to be the expansion of an infinite product of absolutely convergent series, since it is close to be that, but it's not. To get a complete structure of such an expansion, a natural start seems to be: express the reciprocals of the binomial coefficients in terms of the corresponding Beta function integrals (using new integration variables $t_0, t_1,\dots,t_\lambda$ etc). I mean: let's use the identity: $$\frac{1}{{p+q+k \choose k }}=k \frac{(p+q)!(k-1)!}{(p+q+k)!}=k\int_0^1 s^{\\ p+q}(1-s)^{k-1}ds. $$

This way the sum in your last limit, that I will denoe $S_{\lambda+1}(x),$ becomes an integral of a certain product of $\lambda+1$ simpler functions, also depending on the parameter $x$ over $t\in[0,1]^{\lambda+1}.$ (Or, if you like it more, the whole sum $S_\infty$ can be represented as an integral over $[0,1]^{\mathbb{N}}$, with respect to the countable power of the uniform measure, of a certain function $F:\mathbb{C}\times [0,1]^{\mathbb{N}}\to\mathbb{C}$, expressed as infinite product).

Specifically, for $x\in \mathbb{C}$ let's consider the series (essentially, a Bessel function of order 0). $$J(x):=\sum_{n=0}^\infty\frac{ x^{\\, n}}{(n!)^2}.$$ Then one finds: $$S_{\lambda}(x)=\int_{I^{\lambda}}\\,J(xt_0)\\,J(xt_0t_1)\\,J(xt_1t_2)\dots J(xt_{\lambda-2}t_{\lambda-1})\\,J(xt_{\lambda-1})\\ \lambda!\\!\\!\prod_{0\leq k<\lambda}(1-t_j)^{\\,j}\\ dt.$$

(Now, this still needs some work before concluding as you want -the role of the Bessel function here is quite obscure to me; but notice that the latter expression at least gives an account of what's on. We are integrating the product $J(xt_0)J(xt_0t_1)\dots J(xt_{\lambda-2}t_{\lambda-1})J(xt_{\lambda-1})$ with probability measures on the cubes $I^\lambda,$ and these measures are slowly concentrating near $0.$ What makes delicate the evaluation, of course, is that the dimension of the cube is also going to infinity; nevertheless a reasonable guess is that the integral is close to the product of the $J$'s evaluated at the point $x/\lambda$, which would imply $$S_\lambda(x)=\left(1+\frac{x+o(1)}{\lambda}\right)^{\lambda}=\\,e^x+o(1),$$ as $\lambda\to\infty.$)

(edit: I have shifted by one the previous definition of $S_n$ so that now it is a sum over $n$-ples, and equals an $n$-fold iterated integral.)

Example: for $n=5$, the above integral, computed formally with Maple, is $$S_5(x)=1+x+\frac{1}{2}x^2+\frac{1}{6}x^3+\frac{1}{24}x^4+\frac{1}{120}x^5+\frac{1}{720}x^6+\frac{5039}{25401600}x^7+O(x^8).$$

Some hints. The sum indicized on $\mathbb{N}^{< \omega}$ would like to be the expansion of an infinite product of absolutely convergent series, since it is close to be that, but it's not. To get a complete structure of such an expansion, a natural start seems to be: express the reciprocals of the binomial coefficients in terms of the corresponding Beta function integrals (using new integration variables $t_0, t_1,\dots,t_\lambda$ etc). I mean: let's use the identity: $$\frac{1}{{p+q+k \choose k }}=k \frac{(p+q)!(k-1)!}{(p+q+k)!}=k\int_0^1 s^{\\ p+q}(1-s)^{k-1}ds. $$

This way the sum in your last limit, that I will denoe $S_{\lambda+1}(x),$ becomes an integral of a certain product of $\lambda+1$ simpler functions, also depending on the parameter $x$ over $t\in[0,1]^{\lambda+1}.$ (Or, if you like it more, the whole sum $S_\infty$ can be represented as an integral over $[0,1]^{\mathbb{N}}$, with respect to the countable power of the uniform measure, of a certain function $F:\mathbb{C}\times [0,1]^{\mathbb{N}}\to\mathbb{C}$, expressed as infinite product).

Specifically, for $x\in \mathbb{C}$ let's consider the series (essentially, a Bessel function of order 0). $$J(x):=\sum_{n=0}^\infty\frac{ x^{\, n}}{(n!)^2}.$$ Then one finds: $$S_{\lambda}(x)=\int_{I^{\lambda}}\,J(xt_0)\,J(xt_0t_1)\,J(xt_1t_2)\dots J(xt_{\lambda-2}t_{\lambda-1})\,J(xt_{\lambda-1})\ \lambda!\!\!\prod_{0\leq k<\lambda}(1-t_j)^{\,j}\ dt.$$

(Now, this still needs some work before concluding as you want -the role of the Bessel function here is quite obscure to me; but notice that the latter expression at least gives an account of what's on. We are integrating the product $J(xt_0)J(xt_0t_1)\dots J(xt_{\lambda-2}t_{\lambda-1})J(xt_{\lambda-1})$ with probability measures on the cubes $I^\lambda,$ and these measures are slowly concentrating near $0.$ What makes delicate the evaluation, of course, is that the dimension of the cube is also going to infinity; nevertheless a reasonable guess is that the integral is close to the product of the $J$'s evaluated at the point $x/\lambda$, which would imply $$S_\lambda(x)=\left(1+\frac{x+o(1)}{\lambda}\right)^{\lambda}=\,e^x+o(1),$$ as $\lambda\to\infty.$)

(Edit: I have shifted by one the previous definition of $S_n$ so that now it is a sum over $n$-ples, and equals an $n$-fold iterated integral.)

Example: for $n=5$, the above integral, computed formally with Maple, is $$S_5(x)=1+x+\frac{1}{2}x^2+\frac{1}{6}x^3+\frac{1}{24}x^4+\frac{1}{120}x^5+\frac{1}{720}x^6+\frac{5039}{25401600}x^7+O(x^8).$$

Some hints. The sum indicized on $\mathbb{N}^{< \omega}$ would like to be the expansion of an infinite product of absolutely convergent series, since it is close to be that, but it's not. To get a complete structure of such an expansion, a natural start seems to be: express the reciprocals of the binomial coefficients in terms of the corresponding Beta function integrals (using new integration variables $t_0, t_1,\dots,t_\lambda$ etc). I mean: let's use the identity: $$\frac{1}{{p+q+k \choose k }}=k \frac{(p+q)!(k-1)!}{(p+q+k)!}=k\int_0^1 s^{\\ p+q}(1-s)^{k-1}ds. $$

This way the sum in your last limit, that I will denoe $S_{\lambda+1}(x),$ becomes an integral of a certain product of $\lambda+1$ simpler functions, also depending on the parameter $x$ over $t\in[0,1]^{\lambda+1}.$ (Or, if you like it more, the whole sum $S_\infty$ can be represented as an integral over $[0,1]^{\mathbb{N}}$, with respect to the countable power of the uniform measure, of a certain function $F:\mathbb{C}\times [0,1]^{\mathbb{N}}\to\mathbb{C}$, expressed as infinite product).

Specifically, for $x\in \mathbb{C}$ let's consider the series (essentially, a Bessel function of order 0). $$J(x):=\sum_{n=0}^\infty\frac{ x^{\\, n}}{(n!)^2}.$$ Then one finds: $$S_{\lambda}(x)=\int_{I^{\lambda}}\\,J(xt_0)\\,J(xt_0t_1)\\,J(xt_1t_2)\dots J(xt_{\lambda-2}t_{\lambda-1})\\,J(xt_{\lambda-1})\\ \lambda!\\!\\!\prod_{0\leq k<\lambda}(1-t_j)^{\\,j}\\ dt.$$

Now, this still needs some work before concluding as you want (the role of the Bessel function here is quite obscure to me); but notice that the latter expression at least gives an account of what's on. We are integrating the product $J(xt_0)J(xt_0t_1)\dots J(xt_{\lambda-2}t_{\lambda-1})J(xt_{\lambda-1})$ with probability measures on the cubes $I^\lambda,$ and these measures are slowly concentrating near $0.$ What makes delicate the evaluation, of course, is that the dimension of the cube is also going to infinity; nevertheless a reasonable guess is that the integral is close to the product of the $J$'s evaluated at the point $x/\lambda$, which would imply $$S_\lambda(x)=\left(1+\frac{x+o(1)}{\lambda}\right)^{\lambda}=\\,e^x+o(1),$$ as $\lambda\to\infty.$

(edit: I have shifted by one the previous definition of $S_n$ so that now it is a sum over $n$-ples, and equals an $n$-fold iterated integral.)

Example: for $n=5$, the above integral, computed formally with Maple, is $$S_5(x)=1+x+\frac{1}{2}x^2+\frac{1}{6}x^3+\frac{1}{24}x^4+\frac{1}{120}x^5+\frac{1}{720}x^6+\frac{5039}{25401600}x^7+O(x^8).$$

Some hints. The sum indicized on $\mathbb{N}^{< \omega}$ would like to be the expansion of an infinite product of absolutely convergent series, since it is close to be that, but it's not. To get a complete structure of such an expansion, a natural start seems to be: express the reciprocals of the binomial coefficients in terms of the corresponding Beta function integrals (using new integration variables $t_0, t_1,\dots,t_\lambda$ etc). I mean: let's use the identity: $$\frac{1}{{p+q+k \choose k }}=k \frac{(p+q)!(k-1)!}{(p+q+k)!}=k\int_0^1 s^{\\ p+q}(1-s)^{k-1}ds. $$

This way the sum in your last limit, that I will denoe $S_{\lambda+1}(x),$ becomes an integral of a certain product of $\lambda+1$ simpler functions, also depending on the parameter $x$ over $t\in[0,1]^{\lambda+1}.$ (Or, if you like it more, the whole sum $S_\infty$ can be represented as an integral over $[0,1]^{\mathbb{N}}$, with respect to the countable power of the uniform measure, of a certain function $F:\mathbb{C}\times [0,1]^{\mathbb{N}}\to\mathbb{C}$, expressed as infinite product).

Specifically, for $x\in \mathbb{C}$ let's consider the series (essentially, a Bessel function of order 0). $$J(x):=\sum_{n=0}^\infty\frac{ x^{\\, n}}{(n!)^2}.$$ Then one finds: $$S_{\lambda}(x)=\int_{I^{\lambda}}\\,J(xt_0)\\,J(xt_0t_1)\\,J(xt_1t_2)\dots J(xt_{\lambda-2}t_{\lambda-1})\\,J(xt_{\lambda-1})\\ \lambda!\\!\\!\prod_{0\leq k<\lambda}(1-t_j)^{\\,j}\\ dt.$$

(Now, this still needs some work before concluding as you want -the role of the Bessel function here is quite obscure to me; but notice that the latter expression at least gives an account of what's on. We are integrating the product $J(xt_0)J(xt_0t_1)\dots J(xt_{\lambda-2}t_{\lambda-1})J(xt_{\lambda-1})$ with probability measures on the cubes $I^\lambda,$ and these measures are slowly concentrating near $0.$ What makes delicate the evaluation, of course, is that the dimension of the cube is also going to infinity; nevertheless a reasonable guess is that the integral is close to the product of the $J$'s evaluated at the point $x/\lambda$, which would imply $$S_\lambda(x)=\left(1+\frac{x+o(1)}{\lambda}\right)^{\lambda}=\\,e^x+o(1),$$ as $\lambda\to\infty.$)

(edit: I have shifted by one the previous definition of $S_n$ so that now it is a sum over $n$-ples, and equals an $n$-fold iterated integral.)

Example: for $n=5$, the above integral, computed formally with Maple, is $$S_5(x)=1+x+\frac{1}{2}x^2+\frac{1}{6}x^3+\frac{1}{24}x^4+\frac{1}{120}x^5+\frac{1}{720}x^6+\frac{5039}{25401600}x^7+O(x^8).$$

Some hints. The sum indicized on $\mathbb{N}^{< \omega}$ would like to be the expansion of an infinite product of absolutely convergent series, since it is close to be that, but it's not. To get a complete structure of such an expansion, a natural start seems to be: express the reciprocals of the binomial coefficients in terms of the corresponding Beta function integrals (using new integration variables $t_0, t_1,\dots,t_\lambda$ etc). I mean: let's use the identity: $$\frac{1}{{p+q+k \choose k }}=k \frac{(p+q)!(k-1)!}{(p+q+k)!}=k\int_0^1 s^{\\ p+q}(1-s)^{k-1}ds. $$

This way the $\lambda$ sum $S_\lambda(x)$ in your last limit becomes an integral of a certain product of $\lambda+1$ simpler functions, also depending on the parameter $x$ over $t\in[0,1]^{\lambda+1}.$ (Or, if you like it more, the whole sum $S_\infty$ can be represented as an integral over $[0,1]^{\mathbb{N}}$, with respect to the countable power of the uniform measure, of a certain function $F:\mathbb{C}\times [0,1]^{\mathbb{N}}\to\mathbb{C}$, expressed as infinite product).

Specifically, for $x\in \mathbb{C}$ let's consider the series (essentially, a Bessel function of order 0). $$J(x):=\sum_{n=0}^\infty\frac{ x^{\\, n}}{(n!)^2}.$$ Then one finds: $$S_{\lambda}(x)=\int_{I^{\lambda+1}}\\,J(xt_0)\\,J(xt_0t_1)\\,J(xt_1t_2)\dots J(xt_{\lambda-1}t_{\lambda})\\,J(xt_{\lambda})\\ (\lambda+1)!\\!\\!\prod_{0\leq k\leq \lambda}(1-t_k)^k\\ dt.$$

Now, this still needs some work before concluding as you want (the role of the Bessel function here is quite obscure to me); but notice that the latter expression at least gives an account of what's on. We are integrating the product $J(xt_0)J(xt_0t_1)\dots J(xt_{\lambda-1}t_{\lambda})J(xt_{\lambda})$ with probability measures on the cubes $I^{\lambda+1},$ and these measures are slowly concentrating near $0.$ What makes delicate the evaluation, of course, is that the dimension of the cube is also going to infinity; nevertheless a reasonable guess is that the integral is close to the product of the $J$'s evaluated at the point $x/\lambda$, which would imply $$S_\lambda(x)=\left(1+\frac{x+o(1)}{\lambda}\right)^{\lambda}=\\,e^x+o(1),$$ as $\lambda\to\infty.$

Some hints. The sum indicized on $\mathbb{N}^{< \omega}$ would like to be the expansion of an infinite product of absolutely convergent series, since it is close to be that, but it's not. To get a complete structure of such an expansion, a natural start seems to be: express the reciprocals of the binomial coefficients in terms of the corresponding Beta function integrals (using new integration variables $t_0, t_1,\dots,t_\lambda$ etc). I mean: let's use the identity: $$\frac{1}{{p+q+k \choose k }}=k \frac{(p+q)!(k-1)!}{(p+q+k)!}=k\int_0^1 s^{\\ p+q}(1-s)^{k-1}ds. $$

This way the sum in your last limit, that I will denoe $S_{\lambda+1}(x),$ becomes an integral of a certain product of $\lambda+1$ simpler functions, also depending on the parameter $x$ over $t\in[0,1]^{\lambda+1}.$ (Or, if you like it more, the whole sum $S_\infty$ can be represented as an integral over $[0,1]^{\mathbb{N}}$, with respect to the countable power of the uniform measure, of a certain function $F:\mathbb{C}\times [0,1]^{\mathbb{N}}\to\mathbb{C}$, expressed as infinite product).

Specifically, for $x\in \mathbb{C}$ let's consider the series (essentially, a Bessel function of order 0). $$J(x):=\sum_{n=0}^\infty\frac{ x^{\\, n}}{(n!)^2}.$$ Then one finds: $$S_{\lambda}(x)=\int_{I^{\lambda}}\\,J(xt_0)\\,J(xt_0t_1)\\,J(xt_1t_2)\dots J(xt_{\lambda-2}t_{\lambda-1})\\,J(xt_{\lambda-1})\\ \lambda!\\!\\!\prod_{0\leq k<\lambda}(1-t_j)^{\\,j}\\ dt.$$

Now, this still needs some work before concluding as you want (the role of the Bessel function here is quite obscure to me); but notice that the latter expression at least gives an account of what's on. We are integrating the product $J(xt_0)J(xt_0t_1)\dots J(xt_{\lambda-2}t_{\lambda-1})J(xt_{\lambda-1})$ with probability measures on the cubes $I^\lambda,$ and these measures are slowly concentrating near $0.$ What makes delicate the evaluation, of course, is that the dimension of the cube is also going to infinity; nevertheless a reasonable guess is that the integral is close to the product of the $J$'s evaluated at the point $x/\lambda$, which would imply $$S_\lambda(x)=\left(1+\frac{x+o(1)}{\lambda}\right)^{\lambda}=\\,e^x+o(1),$$ as $\lambda\to\infty.$

(edit: I have shifted by one the previous definition of $S_n$ so that now it is a sum over $n$-ples, and equals an $n$-fold iterated integral.)

Example: for $n=5$, the above integral, computed formally with Maple, is $$S_5(x)=1+x+\frac{1}{2}x^2+\frac{1}{6}x^3+\frac{1}{24}x^4+\frac{1}{120}x^5+\frac{1}{720}x^6+\frac{5039}{25401600}x^7+O(x^8).$$