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Limit as n approaches infinity of $$ \left[\left(\frac{2\cdot 1+1}{2\cdot 1+3}\right)^{1^2}\cdot\left(\frac{2\cdot 2+1}{2\cdot 2+3}\right)^{2^2}\cdot \ldots \cdot\left(\frac{2\cdot n+1}{2\cdot n+3}\right)^{n^2}\right]^{1/n}$$

I tried solving it using the sandwich theorem.

I started with the series \begin{align}b_n&=\left[\left(\frac{2\cdot 1+1}{2\cdot 1+3}\right)^{1^2}\cdot \left(\frac{2\cdot 1+1}{2\cdot 1+3}\right)^{2^2}\cdot \ldots \cdot \left(\frac{2\cdot 1+1}{2\cdot 1+3}\right)^{n^2}\right]^{1/n}\\&= \left(\frac{2\cdot 1+1}{2\cdot 1+3}\right)^{(1^2+2^2+3^2*...*n^2)/n}\\&= \left(\frac{2\cdot 1+1}{2\cdot 1+3}\right)^{n*(n+1)*(2*n+1)/n}\end{align} The limit as n approaches infinity of $b_n= 0$

And the series:

\begin{align}c_n&=\left[\left(\frac{2\cdot n+1}{2\cdot n+3}\right)^{1^2}\cdot \left(\frac{2\cdot n+1}{2\cdot n+3}\right)^{2^2}\cdot \ldots \cdot \left(\frac{2\cdot n+1}{2\cdot n+3}\right)^{n^2}\right]^{1/n}\\&= \left(\frac{2\cdot n+1}{2\cdot n+3}\right)^{(1^2+2^2+3^2*...*n^2)/n}\\&= \left(\frac{2\cdot n+1}{2\cdot n+3}\right)^{n*(n+1)*(2*n+1)/n}\end{align}

The limit as $n$ approaches infinity of $c_n=0$

$c_n > a_n > b_n$ and limit as n approaches infinity of $b_n$ = limit as $n$ approaches infinity of $c_n=0$ so from sandwich theorem limit as n approaches infinity of $a_n=0$

Can someone confirm this?

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    $\begingroup$ The MathJax doesn't take effect unless you enclose the expressions in $ signs. $\endgroup$ – saulspatz May 4 at 16:16
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$$ \left[\left(\frac{2\cdot 1+1}{2\cdot 1+3}\right)^{1^2}\cdot\left(\frac{2\cdot2+1}{2\cdot2+3}\right)^{2^2}\cdot \ldots\cdot\left(\frac{2\cdot n+1}{2\cdot n+3}\right)^{n^2}\right]^{1/n}=\left[ \prod_{i=1}^n \left( \frac{2i+1}{2i+3}\right)^{i^2}\right]^\frac{1}{n}$$

A trivial lower bound for the term would be $0$.

Now, let's check for the upper boudn that you proposed:

\begin{align} \left[ \prod_{i=1}^n \left( \frac{2i+1}{2i+3}\right)^{i^2}\right]^\frac{1}{n} & \le \left[ \prod_{i=1}^n \left( \frac{2n+1}{2n+3}\right)^{i^2}\right]^\frac{1}{n}\\ &=\left( \frac{2n+1}{2n+3}\right)^{\frac{n(n+1)(2n+1)}{6n}}\\ &=\left( \frac{2n+1}{2n+3}\right)^{\frac{(n+1)(2n+1)}{6}}\\ &= \exp\left(\frac{(n+1)(2n+1)}6\cdot \ln \left( \frac{2n+1}{2n+3}\right) \right) \end{align}

Taking limit, \begin{align} \lim_{n \to \infty}\exp\left(\frac{(n+1)(2n+1)}6\cdot \ln \left( \frac{2n+1}{2n+3}\right) \right) &=\lim_{n \to \infty}\exp\left(\frac{n^2}3\cdot \ln \left( \frac{2n+1}{2n+3}\right) \right)\\ &=\exp\left(\lim_{n \to \infty}\frac{n^2}3\cdot \ln \left( 1-\frac{2}{2n+3}\right) \right)\\ &=\exp\left(\lim_{n \to \infty}\frac{n^2}{3(n+1.5)}\cdot \ln \left( 1-\frac{1}{n+1.5}\right)^{n+1.5} \right)\\ &=\exp\left(-\lim_{n \to \infty}\frac{n^2}{3(n+1.5)} \right)\\ &=0 \end{align}

Hence the limit is indeed $0$.

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