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$$\lim\limits_{x \to 0}\frac{(n+1)^x +(n+2)^x +....+(2n)^x -n }{x}$$ I tried L'Hospital's Rule, I got $$\lim\limits_{x \to 0} \frac{(n+1)^x ln(n+1) +(n+2)^x ln(n+2) +...+(2n)^x ln(2n) }{1}$$ $$=ln(n+1) +ln(n+2) +.....+ln(2n)$$ $$=ln\left((n+1)(n+2)(n+3)....(2n)\right)$$ but I could not evaluate neither the sum nor the product I want to know how can I evaluate the sum or the product and get a solution without using L'Hospital's Rule if it exists

imranfat
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Hussien Mohamed
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    you can apply Riemann sum – haqnatural Oct 27 '17 at 17:49
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    the result should be $$-\text{log$\Gamma $}(n)+\text{log$\Gamma $}(2 n)+\log (2)$$ – Dr. Sonnhard Graubner Oct 27 '17 at 17:52
  • what is the procedure ? – Hussien Mohamed Oct 27 '17 at 18:05
  • @Dr.SonnhardGraubner I have found a different result. The function is $$\frac{\sum _{k=n}^{2 n} (k+1)^x-n}{x}=\frac{\zeta (-x,n+1)-\zeta (-x,2 n+2)-n}{x}$$ apply l'Hopital rule and plug $x\to 0$ I got $$\log \frac{(2 n)!}{(n-1)!}+\log\frac{2 n+1}{n}$$ – Raffaele Oct 27 '17 at 18:25
  • Your answer is correct and it can't be simplified further. To avoid L'Hospital's Rule just use the standard limit $\lim_{x\to 0}\dfrac {a^{x} - 1}{x}=\log a$. – Paramanand Singh Oct 28 '17 at 09:25
  • How can i eliminate n to get use this form ? @Paramanand Singh – Hussien Mohamed Oct 28 '17 at 12:35
  • Why eliminate $n$? $n$ is a part of question and it will also be in the answer. The expression can be written as $$\sum_{i=1}^{n}\frac{(n+i)^{x}-1}{x}$$ and hence the limit is $$\sum_{i=1}^{n}\log(n+i)=\log\prod_{i=1}^{n}(n+i)$$ same as your answer. – Paramanand Singh Oct 28 '17 at 13:22
  • I am also a bit surprised at the approaches suggested in comments. We don't need Gamma and zeta and even Riemann sum to evaluate such run of the mill simple limits. The method given by the asker itself is one simple approach but it uses L'Hospital's Rule. – Paramanand Singh Oct 28 '17 at 13:37
  • Can we prove that this result is less than $$nln\left(\frac{3n+1}{2}\right)$$ ? – Hussien Mohamed Oct 28 '17 at 21:36

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