Sum of Infinite Series: TI and Book Solutions

[karush]

$\tiny{206.b.46}$
\begin{align*}
\displaystyle
S_{book}&=\sum_{k=1}^{\infty}
\frac{8^k}{k! }=0\\
S_{TI}&=\sum_{k=1}^{\infty}
\frac{8^k}{k! }=e^8-1\\
\end{align*}
$\textsf{ 2 different answers?}$

 

[MarkFL]

I would say your book is wrong. :)

 

[topsquark]

$\tiny{206.b.46}$
\begin{align*}
\displaystyle
S_{book}&=\sum_{k=1}^{\infty}
\frac{8^k}{k! }=0\\
S_{TI}&=\sum_{k=1}^{\infty}
\frac{8^k}{k! }=e^8-1\\
\end{align*}
$\textsf{ 2 different answers?}$

Consider the fact that (8^k)/k! are positive for all k. Thus they cannot sum to 0.

-Dan