Since $\sqrt{1+8n}=\sqrt{8n}\sqrt{1+\frac{1}{8n}}$, and $\frac{1}{8n}<1$ when $n>1$ is an integer, then we can express the real number $\sqrt{1+\frac{1}{8n}}$ by its binomial series. This series starts as $$1+\frac{1}{2}\frac{1}{8n}+\frac{\frac{1}{2}\left(\frac{1}{2}-1\right)}{2!}\left(\frac{1}{8n}\right)^2+ \frac{\frac{1}{2}\left(\frac{1}{2}-1\right)\left(\frac{1}{2}-2\right)}{3!}\left(\frac{1}{8n}\right)^3+\cdots$$
My attempt was thus assuming that $n>1$ is (an even perfect number, it is omitted if you don't read the appendix) a fixed positive integer compute the kth term: the factor $\frac{1}{k!}\left(\frac{1}{8n}\right)^k$ is easiy compute as $\frac{1}{k!2^{3k}n^k}$, and if $k=j+1$, a product $\frac{1}{2}\left(\frac{1}{2}-1\right)\left(\frac{1}{2}-2\right)\cdots\left(\frac{1}{2}-j\right)$ is computed as, since there are $j+1$ factors, $\frac{(-1)^j}{2^{j+1}}\cdot1\cdot 3\cdot 5\cdots(2j-1)$.
My question is
Question. Compute in a closed form the binomial series $(1+\frac{1}{8n})^{1/2}$, where $n>1$ is a fixed integer (thus you show the general term of the expansion in a closed form). Thanks in advance.
Appendix (Optional, I show the context of previous): I've written a post in this Mathematics Stack Exchange that states
Fact 1. If $n$ is an even perfect number then satisfies $$2\sigma(2n)-1-4n=\sqrt{1+8n},$$ where $\sigma(m)$ is the sum of divisors function.
Previous statement is easy to prove using Euler's theorem for even perfect numbers, see [1] for general definitions and statements or this site. Since each even perfect number is a triangle number $n=\frac{q+1}{2}\cdot q=2^{p-1}\cdot(2^p-1)$, where $q=2^p-1$ is the Mersenne's prime corresponding to $n$, it is easy to prove
Fact 2. If $n=\frac{q+1}{2}\cdot q$ is an even perfect number (thus $q=2^p-1$ is its asssociated Mersenne's prime) then $$2q+1=\sqrt{1+8n}$$ (is a positive integer).
My goal is put more thougths on the equation $$2q+1=2\sqrt{q(q+1)}\cdot\left(\text{binomial series corresponding to }\sqrt{1+\frac{1}{8n}}\right),$$ and if it is possible edit new post in this site; I don't know if this way to extract information about even perfect numbers will be useful, suggestions will be welcome. Thanks.
References:
[1] Perfect numbers, Generalized binomial theorem, Sum of divisors function, Mathworld or Wikipedia.
[2] I attempt integrate another factor 2 in the definition of even perfect numbers
By suggestion of an user, see comments, finally we can compute the general term as $$\frac{(-1)^k}{(k+1)2^{5k+4}n^{k+1}}\binom{2k}{k}.$$ Then for every positive integer $$(1+\frac{1}{8n})^{1/2}=1+\sum_{k=0}^\infty \frac{(-1)^k}{(k+1)2^{5k+4}n^{k+1}}\binom{2k}{k}.$$
Too we can made a change in summation index to write for a fixed (and positive) triangle number $n=\frac{t+1}{2}\cdot t$ that $$2\sqrt{t(t+1)}\cdot\left( 1+\sum_{k=1}^\infty \frac{(-1)^{k-1} \binom{2k-2}{k-1}}{k2^{4k-1}t^k(t+1)^k}\right)=2t+1,$$ thus last equation in (myself question) the post it seems that isn't special for even perfect numbers.
Finally, if $n=\frac{q+1}{2}\cdot q$ is an even perfect number, where $q$ is its corresponding Mersenne's prime, then using the equation involving the sum of divisor function, we can write $$2\sigma(2n)-1-4n=2\sqrt{q(q+1)}\cdot\left( 1+\sum_{k=1}^\infty \frac{(-1)^{k-1} \binom{2k-2}{k-1}}{k2^{4k-1}q^k(q+1)^k}\right).$$
It seems that this last, neither is useful.