I would like to evaluate the following sum in terms of $n$. The sum is essentially a weighted sum of a binomial distribution with N = 2n and p = $\frac{1}{n}$. I can't figure out how to do it, despite looking through a lot of posts on here and a few papers. Could anyone help?
$$\sum_{t=1}^{2n} {2n\choose{t}} \left(\frac{1}{n}\right)^t \left(\frac{n-1}{n}\right)^{2n-t} \left(\frac{1}{t}\right)$$
Well this is the key insight for your solution
$$ (1+x)^{2n}=\sum_{r=0}^{2n}{2n\choose r}x^r\\ {(1+x)^{2n}-1\over x}=\sum_{r=1}^{2n}{2n\choose r}x^{r-1} $$
Now we can integrate on both sides from 0 to $1\over n-1$
$$ I=\sum^{2n}_{r=1}{2n\choose r}{1\over r(n-1)^r} $$
Where $$I=\int_0^{1\over n-1}{(1+x)^{2n}-1\over x}dx$$ Now multiply $({n\over n-1})^{2n}$ on both sides to get
$$ \sum_{r=1}^{2n}{2n \choose r}{1\over r}({1\over n})^r({n\over n-1})^{2n-r}=\left({n\over n-1}\right)^{2n}I $$
Closed form of the integral if anyone cares $$ I=2(_3F_2(1,1,1-2n;2,2;-{1\over n})) $$ Where F is the generalised hypergeometric function