MHB What is the sum of this infinite series?

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The infinite series $1 - \frac{2^3}{1!} + \frac{3^3}{2!} - \frac{4^3}{3!} + \cdots$ can be evaluated using the identity $(n+1)^3 = n(n-1)(n-2) + 6n(n-1) + 7n + 1$. By substituting this identity into the series and analyzing the resulting sums, it is determined that terms with negative factorials can be ignored for small values of n. The final evaluation leads to the conclusion that the sum of the series is $-\frac{1}{e}$. This result highlights the application of series manipulation and factorial properties in evaluating infinite series.
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Evaluate the infinite series $1-\dfrac{2^3}{1!}+\dfrac{3^3}{2!}-\dfrac{4^3}{3!}+\cdots$.
 
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anemone said:
Evaluate the infinite series $1-\dfrac{2^3}{1!}+\dfrac{3^3}{2!}-\dfrac{4^3}{3!}+\cdots$.
[sp]
First, you need to check the identity $(n+1)^3 = n(n-1)(n-2) + 6n(n-1) + 7n + 1.$ Then $$ \begin{aligned} \sum_{n=0}^\infty \frac{(-1)^n(n+1)^3}{n!} &= \sum_{n=0}^\infty \frac{(-1)^n \bigl(n(n-1)(n-2) + 6n(n-1) + 7n + 1\bigr)}{n!} \\ &= \sum_{n=0}^\infty (-1)^n \biggl( \frac1{(n-3)!} + 6\frac1{(n-2)!} + 7\frac1{(n-1)!} + \frac1{n!} \biggr). \end{aligned}$$ The last line there looks suspicious, because for small values of $n$ there are some factorials of negative numbers in the denominators. But if you check what happens when $n$ is small, you see that those terms can be ignored. For example, in the first term $$\sum_{n=0}^\infty (-1)^n \frac1{(n-3)!},$$ the summation actually starts at $n=3$ rather than $n=0.$ If you replace the summation index $n$ by $n-3$ then that sum becomes $$\sum_{n=0}^\infty (-1)^{n+3} \frac1{n!} = -\sum_{n=0}^\infty \frac{(-1)^n}{n!}.$$ Therefore $$\sum_{n=0}^\infty \frac{(-1)^n(n+1)^3}{n!} = \sum_{n=0}^\infty \biggl( -\frac{(-1)^n}{n!} + 6\frac{(-1)^n}{n!} - 7\frac{(-1)^n}{n!} + \frac{(-1)^n}{n!} \biggr) = - \sum_{n=0}^\infty\frac{(-1)^n}{n!} = -\frac1e.$$[/sp]
 
Very well done, Opalg and thanks for participating!:)
 
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