MHB Sum of Infinite Series: TI and Book Solutions

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The discussion highlights a discrepancy between two solutions for the infinite series S, with the book stating S = 0 and the TI calculator yielding S = e^8 - 1. Participants argue that the book's solution is incorrect, as the terms (8^k)/k! are positive for all k, making it impossible for the sum to equal zero. The consensus is that the TI solution is accurate, reinforcing the idea that the book's answer is erroneous. This debate underscores the importance of verifying mathematical solutions against established principles.
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
 
There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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