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I'm solving a problem, and I'm getting the following sum as the solution. $$\displaystyle\sum_{k=1}^{n-1}{10k+5\over (k+1)!2^k}$$

Wolfram Alpha says this can be simplified. How do I arrive at the result? I believe I've calculated similar sums in the past, but now I'm completely stumped.

Is there an elementary method to do it? Wolfram Alpha's result doesn't involve any symbols that a high school student wouldn't underdstand so maybe it's doable by elementary means?

Added. I think I made a mistake in obtaining the sum, and it should be $2^{k+1}$ in the denominator instead of $2^k$. I don't know if it changes anything. People were saying something about changing $(k+1)!$ to $k!$ in the comments, so maybe it does, but I completely don't see it. The expression Wolfram Alpha gives now isn't any simpler.

Bartek
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  • Try $10k+5 = 10(k+1) - 5$ and split into two sums in one of which $(k+1)!$ can be replaced by $k!$? – Dilip Sarwate May 20 '13 at 01:17
  • Try a clever form of zero. You have $10k$ in the numerator and a factor of $k+1$ in the denominator. What can you do to kill off the factor of $10k$? – Cameron Williams May 20 '13 at 01:17
  • @CameronWilliams Thank you for the comment. Could you be more explicit about what I should be trying to achieve? I'm not sure what "kill off the factor of $10k$" means, and why I want to do that. – Bartek May 20 '13 at 01:21
  • @DilipSarwate Thank you. Why is $k!$ better than $(k+1)!$? – Bartek May 20 '13 at 01:26
  • One possible technique is to use Gosper's algorithm. – Mhenni Benghorbal May 20 '13 at 01:31
  • @MhenniBenghorbal Thanks. I will look into that, but I don't even know what the word "hypergeometric" means. Isn't there an elementary way to do it? – Bartek May 20 '13 at 01:37
  • @Bartek "Why is $k!$ better than $(k+1)!$? Because the sum $\sum \frac{2^{-k}}{k!}$ is part of the exponential series for $e^{-2}$ and perhaps something might be done with that. – Dilip Sarwate May 20 '13 at 02:36
  • @DilipSarwate Thanks for replying. I was thinking about looking at infinite series, but I don't know what they could have to do with the finite sum other than that the finite sums approximate the infinite sum. What could be a general way to use an infinite series here? – Bartek May 20 '13 at 03:09

1 Answers1

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\begin{align} \sum_{k=1}^{n-1}{\frac{10k+5}{2^k(k+1)!}} & =\sum_{k=1}^{n-1}{\frac{10k+10}{2^k(k+1)!}}-\sum_{k=1}^{n-1}{\frac{5}{2^k(k+1)!}} \\ & =10\sum_{k=1}^{n-1}{\frac{1}{2^kk!}}-10\sum_{k=1}^{n-1}{\frac{1}{2^{k+1}(k+1)!}} \\ & =10\sum_{k=1}^{n-1}{\frac{1}{2^kk!}}-10\sum_{k=2}^{n}{\frac{1}{2^kk!}} \\ & =10(\frac{1}{2^11!}-\frac{1}{2^nn!}) \\ & =5-\frac{10}{2^nn!} \end{align}

Ivan Loh
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  • How do you arrive at the second to last line? – joejacobz May 20 '13 at 05:04
  • @joejacobz \begin{align} & 10\sum_{k=1}^{n-1}{\frac{1}{2^kk!}}-10\sum_{k=2}^{n}{\frac{1}{2^kk!}} \ & =[10(\frac{1}{2^11!})+10\sum_{k=2}^{n-1}{\frac{1}{2^kk!}}]-[10(\frac{1}{2^nn!})+10\sum_{k=2}^{n-1}{\frac{1}{2^kk!}}] \end{align} – Ivan Loh May 20 '13 at 05:07