# A question involving a summation of eigenvalues of the Laplacian operator on $\mathbb{S}^2$

Infinite series involving eigenvalues of the Beltrami-Laplace operator on Riemannian manifolds as well as $$L^p$$-estimates of eigenfunctions arise in the study of the nonlinear Schrödinger equation (NLS) on compact manifolds. Denote by $$\mu_{k} := k(k + 1)$$ the eigenvalue of the operator $$- \Delta_{\mathbb{S}^2}$$ associated to the eigenfunction $$e_k \in C^{\infty}(M)$$. Now, consider the summation $$\sum_{k = 0}^{\infty} \frac{1}{ \langle \mu_k - \alpha \rangle \langle \mu_k \rangle^{\varepsilon}}$$ where $$\alpha > 0$$ (is a positive arbitrary constant), $$\varepsilon > 0$$ and $$\langle x \rangle : = 1 + |x|$$. My question is the following:

$$\langle \mu_k - \alpha \rangle^{-1} \langle \mu_k \rangle^{- \varepsilon} \in \ell^{1}_{k}(\mathbb{N}) \mbox{ },$$

independently of the choice of $$\alpha$$?

My failed attempt was to consider two cases: (Case 1) $$\mu_k \geq 4 \alpha$$. In this case we have $$|\mu_k - \alpha| \geq \frac{3}{4} \mu_k$$ and one can obtain the desired conclusion. (Case 2) $$\mu_k \leq 4 \alpha$$. In this case, we have a finite sum, but I would like to prove that the summation is bounded by a constant which does not depends on $$\alpha$$. Thanks in advance !!!

• What do you mean by "independence in relation to $\alpha$"? Looks to me like you've shown that for any $\alpha \neq \mu_k$, the sequence is summable. Are you trying to draw a stronger conclusion? – Neal Aug 17 '19 at 18:43
• It means that the summation is bounded by a constant which does not depend on \alpha – Marcelo Ng Aug 17 '19 at 18:46
• In that case don't you need to restrict $\alpha$ more? Otherwise seems like you could choose $\alpha$ arbitrarily close to some $\mu_k$ to make the sum arbitrarily large. – Neal Aug 17 '19 at 18:48
• I would like to prove that if $\alpha >0$ is arbitrary, then there exists $C >0$ such that $\sum < C$ where $C$ is not dependent of $\alpha$. – Marcelo Ng Aug 17 '19 at 18:54

Neal pay attention on that $$\langle \mu_k-\alpha\rangle \ge 1$$.
Let $$l\in \mathbb{N}$$ be such that $$\mu_l$$ is closest to $$\alpha$$. Consider $$k=l+i$$, $$i\ge 0$$ and $$k=l-i$$, $$0\le i\le l$$. The rest seems to be easy.