Is $\sum_{k=1}^n{{n}\choose{k}}\left (1+\frac{c}{2n}\right)^{n-k}o\left(\frac{1}{n}\right)^k$ equal to $o(1)$? where $c,n$ are some finite, positive real numbers, and the $o$ is little oh.
My attempt is that $\left (1+\frac{c}{2n}\right)^{n-k}o\left(\frac{1}{n}\right)^k$ is just a constant times $o\left(\frac{1}{n}^k\right)$ and is thus $o\left(\frac{1}{n}\right)^k$, and all
$o\left(\frac1n\right)^K$ terms are dominated by the ${{n}\choose{1}} o\left(\frac1n\right)$ term, which is $o(1)$?
 A: Try binomial theorem
$$S=\sum_{k=1}^{n}\binom{n}{k}\left(1+\frac{c}{2n}\right)^{n-k}o\left(\frac{1}{n}\right)^k=\sum_{k=0}^{n}\binom{n}{k}\left(1+\frac{c}{2n}\right)^{n-k}o\left(\frac{1}{n}\right)^k - \left(1+\frac{c}{2n}\right)^{n}=\\
=\left(1+\frac{c}{2n}+ o\left(\frac{1}{n}\right)\right)^{n} - \left(1+\frac{c}{2n}\right)^{n}=$$
which is
$$=o\left(\frac{1}{n}\right)\left(\sum_{k=0}^{n-1} \left(1+\frac{c}{2n}+ o\left(\frac{1}{n}\right)\right)^{n-k}\left(1+\frac{c}{2n}\right)^{k} \right)\tag{1}$$
From $e^x\geq x+1$ for $\forall x>-1$ we have
$$o\left(\frac{1}{n}\right)\left(\sum_{k=0}^{n-1} 1\right)<S<o\left(\frac{1}{n}\right)\left(\sum_{k=0}^{n-1} e^{\frac{c(n-k)}{2n}+ o\left(\frac{1}{n}\right)(n-k)} \cdot e^{\frac{ck}{2n}}\right)=\\
=o\left(\frac{1}{n}\right)\left(\sum_{k=0}^{n-1} e^{\frac{c}{2}+ o\left(\frac{1}{n}\right)(n-k)}\right)<o\left(\frac{1}{n}\right)\left(\sum_{k=0}^{n-1} e^{\frac{c}{2}+ o\left(1\right)}\right)$$
or
$$n\cdot o\left(\frac{1}{n}\right)<S<n \cdot o\left(\frac{1}{n}\right)e^{\frac{c}{2}+ o\left(1\right)}$$
