Simplify summation with factorial in denominator

Question

I don't understand one of the steps in this solution.

The original problem is:

$E[X^n] = \frac{2^n}{n+1}$. Find the distributions of X.

Solution:

Using the following definition of the moment generating function one can write:

$\psi_X(t) = \sum_{k=0}^\infty \frac{E[X^k]}{k!}t^k= \sum_{k=0}^\infty \frac{\frac{2^k}{k+1}}{k!}t^k = \sum_{k=0}^\infty \frac{2^kt^k}{(k+1)k!} = \sum_{k=0}^\infty \frac{(2t)^k}{(k+1)!}=$

the next step is the one I don't get:

$= \frac{1}{2t} \left( \sum_{k=0}^\infty \frac{(2t)^k}{k!} -1 \right) = \frac{1}{2t}(e^{2t}-1)$

And I know from a previous exercise that this is the moment generating function for $U(0,2)$, hence $X$~$U(0,2)$

Any hints or ideas to the step I don't get? How would one normally go about to expand/simplify such a sum?

Answer 1

After letting $j=k+1$, we get $$\sum_{k=0}^\infty \frac{(2t)^k}{(k+1)!}= \sum_{j=1}^\infty \frac{(2t)^{j-1}}{j!}=\frac{1}{2t}\sum_{j=1}^\infty \frac{(2t)^{j}}{j!} =\frac{1}{2t}\left(\sum_{j=0}^\infty \frac{(2t)^{j}}{j!}-1\right) = \frac{1}{2t}(e^{2t}-1)$$ where $e^x=\sum_{j=0}^{\infty}\frac{x^j}{j!}$ has been used.

Answer 2

Remember the Taylor (or simply, power) series of the exponential function:

$$e^x=\sum_{n=0}^\infty\frac{x^n}{n!}$$

This series has infinite convergence radius, which means it is true for any $\;x\in\Bbb R\;$ . Well, what you have there is just the series for $\;e^{2t}\;$ : just substitute $\;2t=x\;$ in the series... but ... I'm sure that $\;1\;$ in the series must be out of it, i.e.: it should be

$$\frac1{2t}\left(\sum_{n=0}^\infty\frac{(2t)^n}{n!}-1\right)$$