# How to find the solution to a multi-inifite summation?

I'm familiar with techniques to solve a single infinite sum such as:

$$\sum_{i=1}^{\infty} (1 + i) \cdot \frac{1}{2^i}$$

Which ends up being equal to $$3$$, but I'm having trouble figuring out how to tackle a multi summation such as this:

$$\sum_{i=0}^\infty \sum_{j=0}^\infty (3 + i +j) \cdot \left( \frac{1}{3} \right)^{(1+i)} \cdot \left( \frac{2}{3} \right)^{(1+j)}$$

For the first sum I was able to find a pattern in the partial sums. But with this double summation, I don't think that will work as easily. Does anyone know of a technique to solve this second summation analytically? Thanks

A key to calculate this double sum lies in the fact that for absolutely convergent series $$\sum_{i=0}^{\infty}a_i$$ and $$\sum_{j=0}^{\infty}b_j$$ we have

$$\left(\sum_{i=0}^{\infty}a_i\right)\left(\sum_{j=0}^{\infty}b_j\right)=\sum_{i=0}^{\infty}\sum_{j=0}^{\infty}a_ib_j = \sum_{j=0}^{\infty}\sum_{i=0}^{\infty}a_ib_j$$

Now, you can just split up the given series and apply above fact

$$\sum_{i=0}^\infty \sum_{j=0}^\infty (3 + i +j) \left( \frac{1}{3} \right)^{(1+i)} \left( \frac{2}{3} \right)^{(1+j)}$$ $$= \underbrace{3\sum_{i=0}^\infty \sum_{j=0}^\infty \left( \frac{1}{3} \right)^{(1+i)} \left( \frac{2}{3} \right)^{(1+j)}}_{=S_1} + \underbrace{\sum_{i=0}^\infty \sum_{j=0}^\infty i \left( \frac{1}{3} \right)^{(1+i)} \left( \frac{2}{3} \right)^{(1+j)}}_{=S_2} + \underbrace{\sum_{i=0}^\infty \sum_{j=0}^\infty j\left( \frac{1}{3} \right)^{(1+i)} \left( \frac{2}{3} \right)^{(1+j)}}_{=S_3}$$

$$S_1 = 3\cdot \frac 13 \cdot \frac 23 \left(\sum_{i=0}^{\infty}\left( \frac{1}{3} \right)^{i}\right)\left(\sum_{j=0}^{\infty}\left( \frac{2}{3} \right)^{j}\right) = \frac 23\cdot \frac 32 \cdot 3=3$$

$$S_2 = \frac 23 \left(\sum_{i=0}^{\infty}i\left( \frac{1}{3} \right)^{(1+i)}\right)\left(\sum_{j=0}^{\infty}\left( \frac{2}{3} \right)^{j}\right) \stackrel{\sum_{i=0}^{\infty}i x^{i+1} =\frac{x^2}{(1-x)^2}}{=} \frac 23\cdot \frac 14 \cdot 3 = \frac 12$$

$$S_3 = \frac 13 \left(\sum_{i=0}^{\infty}\left( \frac{1}{3} \right)^{i}\right)\left(\sum_{j=0}^{\infty}j\left( \frac{2}{3} \right)^{(1+j)}\right) \stackrel{\sum_{j=0}^{\infty}j x^{j+1} =\frac{x^2}{(1-x)^2}}{=} \frac 13\cdot \frac 32 \cdot 4 = 2$$

So, you get

$$\sum_{i=0}^\infty \sum_{j=0}^\infty (3 + i +j) \left( \frac{1}{3} \right)^{(1+i)} \left( \frac{2}{3} \right)^{(1+j)} = S_1 +S_2 + S_3 = \boxed{\frac{11}{2}}$$

Hint:

$$\sum_{i=0}^\infty \sum_{j=0}^\infty (3 + i +j) \cdot \left( \frac{1}{3} \right)^{(1+i)} \cdot \left( \frac{2}{3} \right)^{(1+j)}$$

$$=\sum_{i=0}^\infty\left( \frac{1}{3} \right)^{(1+i)}\left(\sum_{j=0}^\infty (3 + i +j)\left( \frac{2}{3} \right)^{(1+j)}\right)$$

Now set $$\displaystyle \sum_{j=0}^\infty (3 + i +j)\left( \frac{2}{3} \right)^{(1+j)}=\sum_{j=0}^\infty (3 +j)\left( \frac{2}{3} \right)^{(1+j)}+i\sum_{j=0}^\infty\left( \frac{2}{3} \right)^{(1+j)}$$

We can rewrite the sum as the following, letting $$i+j=k$$:

$$\sum_{k=0}^\infty\sum_{i=0}^k (3+k)(\frac13)^{1+i}(\frac23)^{1+k-j}.$$

Now, the inside sum can be evaluated since it is a geometric series, and the whole sum reduces to something easy to evaluate.