I have the series: $$\sum_{i=1}^{i=10^N} \log_5 i$$ I'm trying to figure out how to get the closed-form solution to this problem. I entered it into WolframAlpha and got that it equals: $ (\log(\mathrm{Pochhammer}(1, 10^N)))/(\log(5))$

I have a few things I'm not understanding about this problem,

  1. What procedure do you use to find a closed-form solution here?
  2. What does this solution even mean? (I've never seen this pochhammer symbol)
  3. If possible, could someone tell me if there's any general method for solving these types of problems? I have a lot of them and have been messing around with some algebra, but don't really even understand how to start them. It seems like I'm just having to google until I find some identity that looks similar enough in form. This is for a programming course, but I have no background in algorithms so I'm struggling a bit.
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    $\begingroup$ $$\sum_{k=1}^K \log(k)=\log(K!)$$ $\endgroup$ – Mark Viola Feb 5 '16 at 23:39
  • $\begingroup$ Does a factorial still count as closed-form? $\endgroup$ – Austin Feb 5 '16 at 23:50
  • 1
    $\begingroup$ You can get a useful approximation by using the Stirling formula. $\endgroup$ – André Nicolas Feb 5 '16 at 23:51

Use $\log_5(x)+\log_5(y)=\log_5(xy)$ to reduce the sum to $$\log_5(10^N!)$$

Here's the definition for Pochhammer symbol ( in this case rising factorial has been used ). $$x^{(n)}:=x(x+1)\cdots(x+n-1)$$ which implies $1^{(n)}=n!$ so our sum is $$\log_5(1^{(10^5)})$$ Wolfram alpha has a built-in function called Pochhammer to evaluate the rising factorial. $$\text{Pochhammer}[x,n]=x^{(n)}$$ So Wolfram alpha interprets our result as $$\log_5(\text{Pochhammer}[1,10^5])$$ But WA prefers base $e$ rather than any arbitrary base so using the identity $\log_ab={\log_c b\over\log_c a}$ for positive numbers $a,b,c;$ WA outputs the final result as $$ {\text{Log}[\text{Pochhammer}[1,10^5]]\over\text{Log[5]}} $$ where Log is the WA function function that calculates logarithm w.r.t $e$ unless we specify otherwise.

  • $\begingroup$ Thanks. I guess there's no straightforward method to solving these besides trying to research different identities? $\endgroup$ – Austin Feb 5 '16 at 23:45
  • $\begingroup$ @Larry Well, if you wrote the summation out, you probably would have noticed you could use logarithm properties to simplify, followed by fancy calculator stuff. $\endgroup$ – Simply Beautiful Art Feb 6 '16 at 1:14

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