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I am trying to find the sum $$\sum _{k=1}^n \frac{k+1}{(k-1)!+k!+(k+1)! }.$$ I tried

Simplify[Sum[(k + 1)/(( k - 1)! + k! + (k + 1)!), {k, 1, n}]]

and got

(-4 - 4 n - n^2 + (1 + n) n! + (1 + n) (1 + n)! + (2 + n)! + n (2 + n)!)/((1 + n) (n! + (1 + n)! + (2 + n)!))

and tried

FullSimplify[Sum[(k + 1)/(( k - 1)! + k! + (k + 1)!), {k, 1, n}]]

1 - 1/Gamma[2 + n]

How can I get the result like Maple?

enter image description here

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  • $\begingroup$ Note that with the default complexity function, definition given as SimplifyCount in the documentation on ComplexityFunction, 1 - 1/Gamma[2 + n] has the same complexity as 1 - 1/Factorial[1 + n] (12). $\endgroup$
    – John Doty
    Commented Aug 23, 2018 at 0:58

2 Answers 2

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Simplify@Simplify`SimplifyGamma[Sum[(k + 1)/((k - 1)! + k! + (k + 1)!), {k, 1, n}]] 
% // TeXForm

$1-\frac{1}{(n+1)!}$

or

Simplify @ Developer`GammaSimplify[Sum[(k + 1)/((k - 1)! + k! + (k + 1)!), {k, 1, n}]] 
% // TeXForm

$1-\frac{1}{(n+1)!}$

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Mathematica's result

1 - 1/Gamma[2 + n]

is equivalent to Maple's for $n \in \mathbb{Z}$, but is also a correct generalization for non-integer $n$.

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