4

Let $$h_{n} = \sum_{k=0}^{n} \dbinom{n}{k} \frac{2^{k+1}}{k+1}.$$ If $$S= \sum_{n=0}^{\infty} \frac{h_{n}}{n!},$$ find $\lfloor S \rfloor $.

I calculated the first few values of n. I got, when n=0, we get 2 when n=1, we get 4 when n=2, we get 26/3

i dont see a pattern

  • https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference see also: https://math.meta.stackexchange.com/questions/9959/how-to-ask-a-good-question/27933#27933 – Integrand Oct 14 '20 at 21:51
  • Ok, that's good. Let's look at the first sum. Aside from the $k+1$ in the denominator, what does it remind you of? – Integrand Oct 14 '20 at 22:06
  • I don't know... –  Oct 14 '20 at 22:07
  • I got it thanks to @Dr.Mathva, thanks for the help Integrand –  Oct 14 '20 at 22:10

2 Answers2

2

Hint. Notice that $$\sum_{k=0}^n \binom{n}{k}\frac{2^{k+1}}{k+1}=\frac{1}{n+1}\cdot \left(3^{n+1}-1\right)$$ Eventhough some combinatorial argument might work in order to prove it, I think that the easiest is considering $\int_0^1 x^k dx = \frac{1}{k+1}$.

In order to finish the problem, simply recall that $$e^x=\sum_{n=0}^\infty\frac{x^n}{n!}$$ I obtained $e^3-e\approx 17.37$

Dr. Mathva
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1

Hint to start:

$$ \eqalign{ & \sum\limits_{\left( {0\, \le } \right)\,\,k\,\,\left( { \le \,n} \right)} {\left( \matrix{ n \hfill \cr k \hfill \cr} \right)x^{\,k} } = \left( {1 + x} \right)^{\,n} \cr & \int_{x = 0}^t {\sum\limits_{\left( {0\, \le } \right)\,\,k\,\,\left( { \le \,n} \right)} {\left( \matrix{ n \hfill \cr k \hfill \cr} \right)x^{\,k} dx} } = \sum\limits_{\left( {0\, \le } \right)\,\,k\,\,\left( { \le \,n} \right)} {\left( \matrix{ n \hfill \cr k \hfill \cr} \right){{t^{\,k + 1} } \over {k + 1}}} = \cr & = \int_{x = 0}^t {\left( {1 + x} \right)^{\,n} dx} = {1 \over {n + 1}}\left( {\left( {1 + t} \right)^{\,n + 1} - 1} \right) \cr} $$

G Cab
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