MHB Sum of Infinite Series: TI and Book Solutions

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karush
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$\tiny{206.b.46}$
\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*}
$\textsf{ 2 different answers?}$
 
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I would say your book is wrong. :)
 
karush said:
$\tiny{206.b.46}$
\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
 

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