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Trying to prove $\lim_{n \to \infty} \frac{1}{n} \left[(n^2 +1^2)(n^2+2^2)^2 \cdots (n^2 + n^2)^n \right]^{\frac{1}{n^2}} = 2e^{-1/2}$

I tried using ratio-root criteria with $a_n= \frac{1}{n^n} \left[(n^2 +1^2)(n^2+2^2)^2 \cdots (n^2 + n^2)^n \right]^{\frac{1}{n}}$ and end up with $\frac{n^n}{(n+1)^{n+1}}\cdot 2(n+1)^2 \Pi_{k=1}^n\frac{\left(k^2\right)^{\frac{k}{n+1}} \left(1+\left(\frac{n+1}{k} \right)^2 \right)^{\frac{k}{n+1}}}{\left(k^2\right)^{\frac{k}{n}} \left(1+\left(\frac{n}{k} \right)^2 \right)^{\frac{k}{n}}}$ and can’t see how to simplify this to get the limit as the numerator and denominator seem to end up going to 1 and some kind of $e^{\Sigma \frac{1}{n}}$ but the $(n+1)^2$ seems to persist and send everything to infinity.

So I tried taking e^ natural log(I’ll denote them as log here, I don’t mind ln either) and got this sum in the exponent $-\log (n) + \Sigma_{k=1}^n \frac{k}{n^2}\log(n^2+k^2)$ comparison showed I could get a $\log 2$ and a $\log n$ But where is the $-1/2$ I need? My guess is that this approach won’t simplify further.

Am doing this for practice/review so would be interested in seeing solutions or hints as I already spent too much time on this. My hunch is that I’m either overlooking something or there’s some typo in the problem.

This is also equivalent to $\lim_{n \to \infty} \frac{1}{n} \left[(n^2 +1^2)^{\frac{1}{n^2}}(n^2+2^2)^{\frac{2}{n^2}} \cdots (n^2 + n^2)^{\frac{1}{n}} \right] = 2e^{-1/2}$

Try explaining this before calling this duplicates. I get something different: $\frac{1}{n} \left[(n^2 +1^2)^{1}(1+\frac{2}{n}^2)^{2} \cdots (n^2 + {n}^2)^{{n}} \right]^{\frac{1}{n^2}} = n^{\frac{1}{n}}\left[(1 +(\frac{1}{n})^2)^{\frac{1}{n^2}}(1+(\frac{2}{n})^2)^{\frac{2}{n^2}} \cdots (1 + (\frac{n}{n})^2)^{\frac{1}{n}} \right]$.

How do you get $\frac{1}{n} \left[(n^2 +1^2)^{1}(1+\frac{2}{n}^2)^{2} \cdots (n^2 + {n}^2)^{{n}} \right]^{\frac{1}{n^2}} = (1 +(\frac{1}{n})^2)(1+(\frac{2}{n})^2)^{2} \cdots (1 + (\frac{n}{n})^2)^{n}$?

user1023185
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2 Answers2

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We have \begin{align*} & \frac{1}{n}\left( {\prod\limits_{k = 1}^n {(n^2 + k^2 )^k } } \right)^{1/n^2 } = \frac{1}{n}\exp \left( {\frac{1}{{n^2 }}\sum\limits_{k = 1}^n {k\log (n^2 + k^2 )} } \right) \\ & = \frac{1}{n}\exp \left( {\frac{1}{n}\sum\limits_{k = 1}^n {\frac{k}{n}\log \left( {1 + \left( {\frac{k}{n}} \right)^2 } \right) + } \frac{{2\log n}}{{n^2 }}\sum\limits_{k = 1}^n k } \right) \\ & = \exp \left( {\frac{1}{n}\sum\limits_{k = 1}^n {\frac{k}{n}\log \left( {1 + \left( {\frac{k}{n}} \right)^2 } \right) + } \frac{{n(n + 1)}}{{n^2 }}\log n - \log n} \right) \\ &= \exp \left( {\frac{1}{n}\sum\limits_{k = 1}^n {\frac{k}{n}\log \left( {1 + \left( {\frac{k}{n}} \right)^2 } \right) + } \frac{{\log n}}{{n }}} \right) \\ & \to \exp \left( {\int_0^1 {x\log (1 + x^2 )dx} +0} \right) = \exp \left( {\log 2 - \frac{1}{2}} \right) = 2 e^{-1/2}. \end{align*}

Gary
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This result for the logarithm that you've got can be calculated further. We have \begin{align} & -\log(n) + \sum_{k=1}^n \frac{k}{n^2} \log(k^2+n^2) = \\ &= -\log(n) + \sum_{k=1}^n \frac{k}{n^2} \Big(2\log n + \log(1+\frac{k^2}{n^2})\Big) = \\ &= -\log(n) + \frac{2\log n}{n^2}\frac{n(n+1)}{2} + \frac{1}{n} \sum_{k=1}^n\frac{k}{n}\log(1+\frac{k^2}{n^2}) = \\ &= \frac{\log n}{n} + \frac{1}{n} \sum_{k=1}^n \frac{k}{n}\log(1+\frac{k^2}{n^2}) \end{align} In the limit $$ \lim_{n\to\infty} (\dots) = 0 + \int_0^1 x\log(1+x^2) dx = \frac12(2\log 2 -1)$$

Adam Latosiński
  • 10,558