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Let $S_n sim N(0,n)$ so that $S_n$ is the partial sum process of standard normal random variables.

My objective is to prove:

$$sum_{n=1}^infty P(S_n geq (1+epsilon)sqrt{2 n log log n }) < infty$$

for all $epsilon > 0$.

I cannot use the law of iterated logarithm, as the reason I want to prove the above inequality is to prove a weaker form of iterated logarithm ($leq 1$ instead of $=1$).

My attempt has been to bound the tail probabilities using known inequalities for standard normal. Firstly, we rewrite it as

$$sum_{n=1}^infty P(S_n/sqrt{n} geq (1+epsilon)sqrt{2 log log n }) < infty$$

$$sum_{n=1}^infty (1-Phi((1+epsilon)sqrt{2 log log n })) < infty$$

Inequalities I know of:

$$1-Phi(x) leq x^{-1} e^{-x^2/2}/sqrt{2pi}$$

$$1-Phi(x) leq frac{1}{2} e^{-x^2/2}$$

But neither is enough to bound the sum by a convergent sequence. Alternatively, I also know that

$1-Phi(x) sim phi(x)$ but that doesn't seem to help

(Too long for a comment)

This cannot be shown in the way you're trying. Indeed, the best possible this way is $Omega(sqrt{log n})$. This is because a commensurate lower bound to the gaussian CDF exists:

$$1 - Phi(x) ge frac{1}{sqrt{2pi}} frac{x}{x^2 + 1} e^{-x^2/2}$$

(to show this, integrate by parts one more time in the proof of the first upper bound you state).

One reason you are failing is the following: with the proof strategy, the following statement would also follow: Consider independent Gaussians $Z_n sim mathcal{N}(0,1).$ Then ${Z_n ge (1+epsilon)sqrt{2log log n} }$ occurs only finitely often a.s. It is intuitive that this is false (for a proof, use the lower bound above and B-C). The LIL truly uses the fact that one is dealing with a sum of Gaussians, by exploting the correlations between $S_n$ and $S_{(1+c)n}$ for small $c$.