Return to Answer

It's a standard theorem in complex analysis that if $z_n$ is a sequence that goes to infinity, there is an entire function taking any prescribed values at the $z_n$. [There is a function $f$ vanishing to order 1 at each $z_n$][1] (for $z_n=n$, you could take $f(z)=\sin \pi z$), and then consider $\sum_n a_nf(z)/(f'(z_n)(z-z_n))$. This may not converge, but you can tweak it by multiplying each term by something that is 1 at $z_n$ (eg, $\exp(c_n(z-z_n))$ for $c_n$ chosen appropriately) to make it converge.

(I don't know off the top of my head how to choose the $c_n$; this is copied from Exercise 1 on page 197 of Ahlfors's Complex Analysis.)

EDIT: It's easy to show that such $c_n$ exist. If you write $b_n=a_n/(z_n f'(z_n))$, then for any fixed $z$, the terms of the sum will be approximately $b_n \exp(c_n z_n)$ for $n$ large. You can obviously pick $c_n$ so that this converges. [1]: http://en.wikipedia.org/wiki/Weierstrass_factorization#Existence_of_entire_function_with_specified_zeroes

It's a standard theorem in complex analysis that if $z_n$ is a sequence that goes to infinity, there is an entire function taking any prescribed values at the $z_n$. There is a function $f$ vanishing to order 1 at each $z_n$ (for $z_n=n$, you could take $f(z)=\sin \pi z$), and then consider $\sum_n a_nf(z)/(f'(z_n)(z-z_n))$. This may not converge, but you can tweak it by multiplying each term by something that is 1 at $z_n$ (eg, $\exp(c_n(z-z_n))$ for $c_n$ chosen appropriately) to make it converge.

(I don't know off the top of my head how to choose the $c_n$; this is copied from Exercise 1 on page 197 of Ahlfors's Complex Analysis.)

EDIT: It's easy to show that such $c_n$ exist. If you write $b_n=a_n/(z_n f'(z_n))$, then for any fixed $z$, the terms of the sum will be approximately $b_n \exp(c_n z_n)$ for $n$ large. You can obviously pick $c_n$ so that this converges.

It's a standard theorem in complex analysis that if z_n is a sequence that goes to infinity, there is an entire function taking any prescribed values at the z_n. [There is a function f vanishing to order 1 at each z_n][1] (for z_n=n, you could take f(z)=\sin \pi z), and then consider \sum_n a_nf(z)/(f'(z_n)(z-z_n)). This may not converge, but you can tweak it by multiplying each term by something that is 1 at z_n (eg, exp(c_n(z-z_n)) for c_n chosen appropriately) to make it converge.

(I don't know off the top of my head how to choose the c_n; this is copied from Exercise 1 on page 197 of Ahlfors's Complex Analysis.)

EDIT: It's easy to show that such c_n exist. If you write b_n=a_n/(z_n f'(z_n)), then for any fixed z, the terms of the sum will be approximately b_n exp(c_n z_n) for n large. You can obviously pick c_n so that this converges. [1]: http://en.wikipedia.org/wiki/Weierstrass_factorization#Existence_of_entire_function_with_specified_zeroes

It's a standard theorem in complex analysis that if $z_n$ is a sequence that goes to infinity, there is an entire function taking any prescribed values at the $z_n$. [There is a function $f$ vanishing to order 1 at each $z_n$][1] (for $z_n=n$, you could take $f(z)=\sin \pi z$), and then consider $\sum_n a_nf(z)/(f'(z_n)(z-z_n))$. This may not converge, but you can tweak it by multiplying each term by something that is 1 at $z_n$ (eg, $\exp(c_n(z-z_n))$ for $c_n$ chosen appropriately) to make it converge.

(I don't know off the top of my head how to choose the $c_n$; this is copied from Exercise 1 on page 197 of Ahlfors's Complex Analysis.)

EDIT: It's easy to show that such $c_n$ exist. If you write $b_n=a_n/(z_n f'(z_n))$, then for any fixed $z$, the terms of the sum will be approximately $b_n \exp(c_n z_n)$ for $n$ large. You can obviously pick $c_n$ so that this converges. [1]: http://en.wikipedia.org/wiki/Weierstrass_factorization#Existence_of_entire_function_with_specified_zeroes

It's a standard theorem in complex analysis that if z_n is a sequence that goes to infinity, there is an entire function taking any prescribed values at the z_n. There is a function f vanishing to order 1 at each z_n (for z_n=n, you could take f(z)=\sin \pi z), and then consider \sum_n a_nf(z)/(f'(z_n)(z-z_n)). This may not converge, but you can tweak it by multiplying each term by something that is 1 at z_n (eg, exp(c_n(z-z_n)) for c_n chosen appropriately) to make it converge.

(I don't know off the top of my head how to choose the c_n; this is copied from Exercise 1 on page 197 of Ahlfors's Complex Analysis.)

It's a standard theorem in complex analysis that if z_n is a sequence that goes to infinity, there is an entire function taking any prescribed values at the z_n. [There is a function f vanishing to order 1 at each z_n][1] (for z_n=n, you could take f(z)=\sin \pi z), and then consider \sum_n a_nf(z)/(f'(z_n)(z-z_n)). This may not converge, but you can tweak it by multiplying each term by something that is 1 at z_n (eg, exp(c_n(z-z_n)) for c_n chosen appropriately) to make it converge.

(I don't know off the top of my head how to choose the c_n; this is copied from Exercise 1 on page 197 of Ahlfors's Complex Analysis.)

EDIT: It's easy to show that such c_n exist. If you write b_n=a_n/(z_n f'(z_n)), then for any fixed z, the terms of the sum will be approximately b_n exp(c_n z_n) for n large. You can obviously pick c_n so that this converges. [1]: http://en.wikipedia.org/wiki/Weierstrass_factorization#Existence_of_entire_function_with_specified_zeroes