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Question Number 215748 by MrGaster last updated on 17/Jan/25

$$\underset{n\to \mathrm{\infty }}{\lim }{\left[\underset{k=1}{\prod ^{n+1}}\mathrm{\Gamma }\left(\frac{1}{k}\right)\right]}^{\frac{1}{n+1}}-{\left[\underset{k=1}{\prod ^n}\mathrm{\Gamma }\left(\frac{1}{k}\right)\right]}^{\frac{1}{n}}$$

Answered by MrGaster last updated on 02/Feb/25

$$=\underset{n\to \mathrm{\infty }}{\lim }{\left[\underset{k=1}{\prod ^{n+1}}\mathrm{\Gamma }\left(\frac{1}{k}\right)\right]}^{\frac{1}{n+1}}[1-{\left(\frac{\underset{k=1}{\prod ^n}\mathrm{\Gamma }\left(\frac{1}{k}\right)}{\underset{k=1}{\prod ^{n+1}}\mathrm{\Gamma }\left(\frac{1}{k}\right)}\right)}^{\frac{1}{n}}]$$$$=\underset{n\to \mathrm{\infty }}{\lim }{\left[\underset{k=1}{\prod ^{n+1}}\mathrm{\Gamma }\left(\frac{1}{k}\right)\right]}^{\frac{1}{n+1}}[1-{\left(\frac{\mathrm{\Gamma }\left(\frac{1}{n+1}\right)}{\underset{k=1}{\prod ^{n+1}}\mathrm{\Gamma }\left(\frac{1}{k}\right)}\right)}^{\frac{1}{n}}]$$$$=\underset{n\to \mathrm{\infty }}{\lim }{\left[\underset{k=1}{\prod ^{n+1}}\mathrm{\Gamma }\left(\frac{1}{k}\right)\right]}^{\frac{1}{n+1}}[1-{\left(\frac{\mathrm{\Gamma }\left(\frac{1}{n+1}\right)}{\underset{k=1}{\prod ^{n+1}}\mathrm{\Gamma }\left(\frac{1}{k}\right)}\right)}^{\frac{1}{n}}]$$$${\left[\underset{k=1}{\prod ^{n+1}}\mathrm{\Gamma }\left(\frac{1}{k}\right)\right]}^{\frac{1}{n+1}}[1-{\left(\frac{\mathrm{\Gamma }\left(\frac{1}{n+1}\right)}{\mathrm{\Gamma }(1+\frac{1}{n+1})}\right)}^{\frac{1}{n}}]$$$$=\underset{n\to \mathrm{\infty }}{\lim }{\left[\underset{k=1}{\prod ^{n+1}}\mathrm{\Gamma }\left(\frac{1}{k}\right)\right]}^{\frac{1}{n+1}}[1-{\left(\frac{1}{\frac{1}{n+1}}\right)}^{\frac{1}{n}}]$$$$=\underset{n\to \mathrm{\infty }}{\lim }{\left[\underset{k=1}{\prod ^{n+1}}\mathrm{\Gamma }\left(\frac{1}{k}\right)\right]}^{\frac{1}{n+1}}[1-{(n+1)}^{\frac{1}{n}}]$$$$=\underset{n\to \mathrm{\infty }}{\lim }{\left[\underset{k=1}{\prod ^{n+1}}\mathrm{\Gamma }\left(\frac{1}{k}\right)\right]}^{\frac{1}{n+1}}[1-e^{\frac{\ln (n+1)}{n}}]$$$$=\underset{n\to \mathrm{\infty }}{\lim }{\left[\underset{k=1}{\prod ^{n+1}}\mathrm{\Gamma }\left(\frac{1}{k}\right)\right]}^{\frac{1}{n+1}}[1-(1+\frac{\ln (n+1)}{n}+O\left(\frac{{\ln }^2(n+1)}{n}\right))]$$$$=\underset{n\to \mathrm{\infty }}{\lim }{\left[\underset{k=1}{\prod ^{n+1}}\mathrm{\Gamma }\left(\frac{1}{k}\right)\right]}^{\frac{1}{n+1}}[-\frac{\ln (n+1)}{n}+O\left(\frac{{\ln }^2(n+1)}{n^2}\right)]$$$$=\underset{n\to \mathrm{\infty }}{\lim }{\left[\underset{k=1}{\prod ^{n+1}}\mathrm{\Gamma }\left(\frac{1}{k}\right)\right]}^{\frac{1}{n+1}}[-\frac{\ln (n-1)}{n}]$$$$=\underset{n\to \mathrm{\infty }}{\lim }{\left[\underset{k=1}{\prod ^{n+1}}\mathrm{\Gamma }\left(\frac{1}{k}\right)\right]}^{\frac{1}{n+1}}[-\frac{1}{n}\ln (n+1)]$$$$=0$$$$$$

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