# How do I make sums full size in cases?

• May 15th 2010, 02:57 AM
scorpion007
How do I make sums full size in cases?
I.e. in the following, the sums are inline, but I'd like them larger.

$\displaystyle f(z)=\begin{cases}-\frac{1}{z-2}\sum_{n=0}^\infty \left(\frac{z-2}{2}\right)^2, & \text{for}~ |z-2|<2 \\ \frac{2}{(z-2)^2} \sum_{n=0}^\infty \left(-\frac{2}{z-2}\right)^2, &\text{for}~|z-2|>2 \end{cases}$

Is it possible?
• May 15th 2010, 03:11 AM
Failure
Quote:

Originally Posted by scorpion007
I.e. in the following, the sums are inline, but I'd like them larger.

$\displaystyle f(z)=\begin{cases}-\frac{1}{z-2}\sum_{n=0}^\infty \left(\frac{z-2}{2}\right)^2, & \text{for}~ |z-2|<2 \\ \frac{2}{(z-2)^2} \sum_{n=0}^\infty \left(-\frac{2}{z-2}\right)^2, &\text{for}~|z-2|>2 \end{cases}$

Is it possible?

If you add \displaystyle to the two cases, you get this:

$\displaystyle f(z)=\begin{cases}\displaystyle-\frac{1}{z-2}\sum_{n=0}^\infty \left(\frac{z-2}{2}\right)^2, & \text{for}~ |z-2|<2 \\ \displaystyle\frac{2}{(z-2)^2} \sum_{n=0}^\infty \left(-\frac{2}{z-2}\right)^2, &\text{for}~|z-2|>2 \end{cases}$
• May 15th 2010, 03:17 AM
scorpion007
Thanks a lot!