The Easy Peasy Double Sum of Series

After switching the order of summation and expanding, I get $\sum_{n=0}^{2014} \frac{(2/3)^{n+1}}{n!} \sum_{m=0}^\infty m(m-1)(\cdots)(m-n+1) (2/3)^{m-n}$ .

This inner series is the $n$ th derivative of $\sum x^m$ , evaluated at $x = 2/3$ . The $n$ th derivative of $\frac1{1-x}$ is $\frac{n!}{(1-x)^{n+1}}$ , so the inner series is $3^{n+1} n!$ .

Putting it together, we get $\sum_{n=0}^{2014} 2^{n+1} = 2^{2016} - 2$ , so the answer is $2016+1+0 = \fbox{2017}$ .

I will use complex analysis to solve it.

$\displaystyle \sum_{m=0}^{\infty}{\sum_{n=0}^{2014}{\binom{m}{n}{\left(\frac{2}{3}\right)}^{m+1}}}$

We can write $\displaystyle \binom{m}{n} = \frac{1}{2\pi\iota}\oint{\frac{{(1+z)}^{m}}{{z}^{n+1}}}$

Well, the above is true because of the fact that when we will find the residue - remember, that they are just Laurent series expansion's "analytic value". So, it will be ${(1+x)}^{m}$ as our analytic function and its series expansion's nth value will be our answer as what we have got.

We will not write the $\frac{1}{2\pi\iota}$ repeatedly. As we know that it is going to "cut out" after we've taken the residues at the end.

$\displaystyle \oint{\sum_{m=0}^{\infty}{\sum_{n=0}^{2014}{\frac{{(1+z)}^{m}}{{z}^{n+1}}{\left(\frac{2}{3}\right)}^{m+1}}}}$

Now, we can use our all famous GP sum formula and simply get,

$\displaystyle 2 \oint{\frac{{z}^{2015}-1}{(1-2z)(z-1)({z}^{2015})}}$

Now, we will use our residue theorem but around what curve? Well, we need to take a curve for which $\left | \frac{2(1+z)}{3} \right | \le 1$

So, $|z| = \frac{1}{2}$ would do all good.

Now, we would take residues but remember we'll omit the $\frac{1}{2\pi\iota}$

Our sum becomes -

$\displaystyle 2\left(\text{Res}\left(\frac{{z}^{2015}-1}{(1-2z)(z-1)({z}^{2015})}, 0\right) + \text{Res}\left(\frac{{z}^{2015}-1}{(1-2z)(z-1)({z}^{2015})}, 1/2\right)\right)$

We can see that the first summand eventually becomes 0. And the 2nd becomes

$\displaystyle 2({2}^{2015} - 1) = {2}^{2016} - 2$

This is because the leading coefficient was negative for the $1-2x$ term and so we changed the sign.