# How to determine $\Sigma_{n=1}^{\infty}\frac{2^n-1}{n!}$ converges?

Given $$A = \begin{bmatrix} 1 & 1 \\ & 2 \end{bmatrix}$$,

$$e^A$$ is computed by the formula

$$e^A= \begin{bmatrix} 1 & \\ & 1 \end{bmatrix} + \frac{1}{1!}\begin{bmatrix} 1 & 1 \\ & 2 \end{bmatrix} + \frac{1}{2!}\begin{bmatrix} 1 & 3 \\ & 4 \end{bmatrix} + \dots = \begin{bmatrix} e & * \\ & e^2 \end{bmatrix}$$.

I am trying to figure out what $$*$$ is in the above formula.

By calculating $$A^n$$,

$$A=\begin{bmatrix} 1 & 1 \\ & 2 \end{bmatrix}$$,

$$A^2=\begin{bmatrix} 1 & 1 \\ & 2 \end{bmatrix}\begin{bmatrix} 1 & 1 \\ & 2 \end{bmatrix}=\begin{bmatrix} 1 & 1+2 \\ & 2^2 \end{bmatrix}$$,

$$A^n=\begin{bmatrix} 1 & 1+2+2^2+\dots+2^{n-1} \\ & 2^n \end{bmatrix}=\begin{bmatrix} 1 & \frac{2^n-1}{2-1} \\ & 2^n \end{bmatrix}$$.

Thus $$* = \Sigma_{n=1}^{\infty}\frac{2^n-1}{n!}$$.

My question is (1) whether I came to the right place until now, (2) if the sum $$*$$ exists and (3) if so, what it is(does it have a explicit formula).

• For series look to exponent en.wikipedia.org/wiki/Taylor_series#Exponential_function – zkutch Jul 17 '20 at 11:24
• What is the lower left entry? Zero? Same as upper entry? – Oscar Lanzi Jul 17 '20 at 11:30
• @OscarLanzi Zero it is. – Henry Choi Jul 17 '20 at 11:32
• But $A^0\ne\begin{bmatrix} 1 & 1 \\ & 1 \end{bmatrix}$, and $A^2\ne\begin{bmatrix} 1 & 3 \\ & 3 \end{bmatrix}$. – TonyK Jul 17 '20 at 11:54
• And finally, $* = \Sigma_{n=1}^{\infty}\frac{2^n-1}{n!}$, not $1+\Sigma_{n=1}^{\infty}\frac{2^n-1}{n!}$. I think I'm done now. – TonyK Jul 17 '20 at 12:03

You might want to use the fact that $$e^x = \sum_{n\geq 0} \dfrac{x^n}{n!}.$$ Hence $$\sum_{n=1}^{\infty} \dfrac{2^n - 1}{n!} = \sum_{n\geq 0} \dfrac{2^n - 1^n}{n!} = e^2 - e^1 = e^2 - e.$$

It does converge, for example by the ratio test.

You can find the limit by splitting the sum into two: it's $$\sum \frac{2^n}{n!} - \sum \frac{1}{n!}$$. Both of these you can recognise from $$e^x = \sum \frac{x^n}{n!}$$.

$$\Sigma_{n=1}^{\infty}\frac{x^n}{n!}$$ is a powerseries with radius of convergence $$R=+\infty$$. Thus both $$\Sigma_{n=1}^{\infty}\frac{2^n}{n!}$$ and $$\Sigma_{n=1}^{\infty}\frac{-1}{n!}=-\Sigma_{n=1}^{\infty}\frac{1}{n!}=-\Sigma_{n=1}^{\infty}\frac{1^n}{n!}$$ are convergent. Take the sum of two.

By the ratio test

$$\frac{2-2^{-n}}{n+1}\le\frac34$$ as soon as $$n>1$$.

(Of course, you can split the two terms and use the well-known Taylor development.)