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It had been shown that $$\sum_{n=0}^{\infty}\frac{(n+1)^2}{(2n+1)!}=\frac{1}{4}(2e+\cosh(1)).$$

Later on, I calculated that $$\sum_{n=0}^{\infty}\frac{(n+1)^3}{(2n+1)!}=\frac{1}{8}(7e+\cosh(1))$$ $$\sum_{n=0}^{\infty}\frac{(n+1)^4}{(2n+1)!}=\frac{1}{16}(25e+2\sinh(1))$$ $$\sum_{n=0}^{\infty}\frac{(n+1)^5}{(2n+1)!}=\frac{1}{32}(97e+9\sinh(1))$$ $$\sum_{n=0}^{\infty}\frac{(n+1)^6}{(2n+1)!}=\frac{1}{64}(434e+9\sinh(1))$$

I used Ahmed S. Attaalla's method, and for more detail please visit How to evaluate $1+\frac{2^2}{3!}+\frac{3^2}{5!}+\frac{4^2}{7!}+\cdots$.

Are numbers such as $7,25,97,434$ related to the Bell's Number? Is here a pattern or general formula for $$\sum_{n=0}^{\infty}\frac{(n+1)^k}{(2n+1)!}$$

Larry
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1 Answers1

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Define $$A_k=\sum_{n=0}^\infty \frac{(n+1)^k}{(2n+1)!}$$ and $$A(t)=\sum_{k=0}^\infty\frac{A_k}{k!}t^k.$$ Then \begin{align} A(t)&=\sum_{n=0}^\infty\frac1{(2n+1)!}\sum_{k=0}^\infty\frac{(n+1)^k}{k!}t^k =\sum_{n=0}^\infty\frac{e^{(n+1)t}}{(2n+1)!}\\ &=e^{t/2}\sum_{n=0}^\infty\frac{e^{(2n+1)t/2}}{(2n+1)!} =e^{t/2}\sinh(e^{t/2}) =e^{t/2}\frac{\exp(e^{t/2})-\exp(-e^{{t/2}})}2. \end{align}

From a classical formula, $$\exp(e^t)=e\sum_{k=0}^\infty B_k\frac{t^k}{k!}$$ with $B_k$ the $k$-th Bell number. A less well-known formula is $$\exp(-e^t)=e^{-1}\sum_{k=0}^\infty R_k\frac{t^k}{k!}$$ with $R_k$ the $k$-th Rao Uppuluri-Carpenter number.

Then we eventually get $$A_k=\frac{e}{2^k}\sum_{0\le r<k/2}{k\choose 2r+1}B_{2r+1} -\frac{e^{-1}}{2^k}\sum_{0\le r<k/2}{k\choose 2r+1}R_{2r+1}.$$

Angina Seng
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  • Can explain it how derived last formula A[k]? Thanks. – Mariusz Iwaniuk Sep 16 '18 at 13:59
  • It's just extracting the coefficient of $t^k/k!$ in the formula got for $A(t)$. – Angina Seng Sep 16 '18 at 14:54
  • In Mathematica I tired A[k_] := E/2^k* Sum[Binomial[k, 2 r + 1]*BellB[2 r + 1], {r, 0, k/2}] - 1/(E*2^k)*Sum[Binomial[k, 2 r + 1]*BellB[2 r + 1, -1], {r, 0, k/2}];A[2] // N(*1.5430806348152437*) ,but correct answer is for k=2-> 1.74491 ? – Mariusz Iwaniuk Sep 16 '18 at 15:00