How can I find a finite closed form for $\phi (\lambda)$?

Question

How can I find a finite closed form for $\phi (\lambda)$, which that

$$\lim_{\lambda \to \infty} \frac{\sum_{n=1}^{\lambda}2^{n^2}}{\phi(\lambda)}=1~~~~~~~~\lambda \in \mathbb{Z^{+}}$$

Is this mathematically possible?

Answer 1

Because $2^{n^2}$ increases so quickly, the sum is essentially just the last term and you can take $\phi(\lambda)=2^{\lambda^2}$. In more detail, $$\frac{\sum_{n=1}^\lambda 2^{n^2}}{2^{\lambda^2}} =1+\frac{\sum_{n=1}^{\lambda-1} 2^{n^2}}{2^{\lambda^2}} \le1+\frac{\lambda2^{(\lambda-1)^2}}{2^{\lambda^2}} =1+\frac{2\lambda}{2^{2\lambda}}$$ which tends to $1$ as $\lambda\to\infty$. The left hand side is clearly at least $1$, so by the squeeze theorem it also tends to $1$ as $\lambda\to\infty$.