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We have the inequality $$\alpha_n(t) \le 4\pi(3\pi)^{1/3} \exp\left\{\int_0^t(1+3F_P(\sigma)) \, d\sigma \right\} \cdot \int_0^t P(\sigma)^2(3CD_1^2)^{1/3}\alpha_{n-1}(\sigma) \, d\sigma$$ for $n=2,3,\ldots$. (We notice that $\alpha_n$ appears on both sides of the inequality.)

Why does it follow that the infinite series $$\sum_{n=1}^\infty \alpha_n(t)$$ converges locally uniformly on $\mathbb R_0^+$ (that is, converges on $[0,T]$ for $0 \le t \le T$)?

This comes from page 354 of the journal that contains the paper "Global symmetric solutions of the initial value problem of stellar dynamics" by Jurgen Batt.

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Some estimates like $\alpha_n(s)\leqslant A(T)^n/n!$ for all $s\in [0,T]$ should hold by induction, if we suppose something moderate on your functions, which depend on $\sigma$. This yields desired uniform convergence.

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  • $\begingroup$ As soon as we bounded $\alpha_n(t)$ by $M_n := A(T)^n/n!$ for all $0 \le t \le T$, we invoked the Weierstrass $M$-test to assert local uniform convergence of the infinite series, right? (I know, basic question, sorry) $\endgroup$ – cupcake Apr 20 '16 at 22:11
  • $\begingroup$ Yes, Weiersteass test, exactly. $\endgroup$ – Fedor Petrov Apr 20 '16 at 22:16

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