I'm quoting S.Ross in his book;
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By using an approximation, due to Stirling, which asserts that $ n! ∼ (n/e)^n \sqrt{2πn} $ where we say that $a_n ∼ b_n$ when $ \lim_{n \to \infty}a_n/b_n = 1 $, we obtain
$P_{00}^{2n} ∼ \frac{(4p(1 − p))^n}{ \sqrt{ \pi n}} $
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I have have derived this myself, and it makes sense, but he then says:
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Now it is easy to verify, for positive $a_n, b_n $, that if $a_n ∼ b_n$, then $ \sum_n a_n < \infty $ if and only if $ \sum_n b_n < \infty $. $ \star$
Hence, $ \sum_{n=1}^\infty {P_{00}^{2n}}$ will converge if and only if $ \sum_{n=1}^\infty { \frac{(4p(1 − p))^n}{ \sqrt{ \pi n}}}$ does. $ \star \star$
However, $ 4p(1 − p) \leq 1$ with equality holding if and only if $p = 1/2$. Hence, $ \sum_{n=1}^\infty {P_{00}^{2n}}=\infty$ if and only if $ p = 1/2 $.
Thus, the chain is recurrent when $ p = 1/2 $ and transient if $p \neq 1/2$.
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First of all, I'm not sure what he means in $ \star$. Is the equivalence relation that loose? Any fraction of sums with values sufficiently large enough to pass as "less than infinity" equals 1?
Secondly, putting in for $p=1/2$ does the second sum in $ \star \star$ really converge to infinity? In other words, with the numerator equal to 1, can you tell me why $ \sum_{n=1}^\infty { \frac{1}{ \sqrt{ \pi n}}} = \infty $ ?
I understand that with $p=1/2$ in a symmetric random walk, state 0 will be revisited infinitely many times, but I just can't get this mathematical explanation of the concept.
Anyone who can answer the two points I made?
Regarding $\star$, if $a_n \sim b_n$, then we know that there exists an $N$ such that for $n \ge N$, $\frac12 < \frac{a_n}{b_n} < 2$. This tells us that if $\sum_{n \ge N} a_n$ converges, then $\sum_{n \ge N} b_n < \sum_{n \ge N} 2a_n$ converges, and vice versa; the initial segments $a_1 + a_2 + \dots + a_{N-1}$ and $b_1 + b_2 + \dots + b_{N-1}$ don't affect the convergence.
As for $\star\star$ when $p=\frac12$, $\sum_{n\ge1} \frac1{\sqrt n}$ diverges either by the integral test (the integral of $\frac1{\sqrt x}$ from $1$ to $\infty$ diverges), or by comparing it to $\sum_{n\ge1} \frac1n$ and remembering that the harmonic series diverges.