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I wish to determine the limit of $a_n$, where we recursively define:

$$a_n= a_{n-1} \left(1+ \frac{1}{\sqrt n}\right)$$ where $a_0=1$

I already noticed it is increasing, because $$a_n - a_{n-1}= \frac{1}{\sqrt{n}}a_{n-1}>0 $$ Since all terms are strictly positive. (alternatively, we could form a better argument via induction).


edit:

Base case: $a_1 = 2 > 1 = a_0$.

Hypothesis: $a_k > a_{k-1}$

Step: $a _ {k+1}= a_{k} \left(1+ \frac{1}{\sqrt n}\right)>a_{k-1} \left(1+ \frac{1}{\sqrt n}\right)=a_k $ Therefore $a_n >a_{n-1}$ for all $n$ and the sequence is increasing.


I plotted some terms and this seems to grow exponentially, how would I prove it diverges? I think I need to show it is unbounded.I tried the ratio test for sequences but this is inconclusive as the ratio tends to $1$.

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    $\begingroup$ Think about $\log a_n$. $\endgroup$ – Lord Shark the Unknown Jan 27 at 13:03
  • $\begingroup$ It doesn't look very exponential from this angle... $\endgroup$ – Rhys Jan 27 at 13:05
  • $\begingroup$ wolframalpha.com/input/… $\endgroup$ – Wesley Strik Jan 27 at 13:05
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    $\begingroup$ You’re asking whether the infinite product $\prod(1+1/\sqrt n)$ converges. This question should be closely relatable to the convergence of $\sum1/\sqrt n$. $\endgroup$ – Lubin Jan 27 at 13:17
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    $\begingroup$ Hint $$a_n = a_{n-1}\left(1+\frac{1}{\sqrt{n}}\right)=a_{n-2}\left(1+\frac{1}{\sqrt{n-1}}\right)\left(1+\frac{1}{\sqrt{n}}\right)=\\ a_0\prod\limits_{k=1}^n\left(1+\frac{1}{\sqrt{k}}\right)\geq \left(1+\frac{1}{\sqrt{n}}\right)^n= \left(\left(1+\frac{1}{\sqrt{n}}\right)^{\sqrt{n}}\right)^{\sqrt{n}} \approx e^{\sqrt{n}}$$ $\endgroup$ – rtybase Jan 27 at 13:31
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Further to my comment, starting with $$a_n = a_{n-1}\left(1+\frac{1}{\sqrt{n}}\right)=a_{n-2}\left(1+\frac{1}{\sqrt{n-1}}\right)\left(1+\frac{1}{\sqrt{n}}\right)=\\ a_0\prod\limits_{k=1}^n\left(1+\frac{1}{\sqrt{k}}\right)\geq \left(1+\frac{1}{\sqrt{n}}\right)^n= \left(\left(1+\frac{1}{\sqrt{n}}\right)^{\sqrt{n}}\right)^{\sqrt{n}} \geq ...$$ and applying Bernoulli's inequality $$...\geq \left(1+\frac{\sqrt{n}}{\sqrt{n}}\right)^{\sqrt{n}}=2^{\sqrt{n}}$$ As a result $$a_n \geq 2^{\sqrt{n}}$$ and the result follows, i.e. the sequence is unbounded and diverges.

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  • $\begingroup$ So the trick is to estimate every term by the smallest term and show that we still get divergence either by Bernoulli's inequality or by comparison with $e^{\sqrt{n}}$ $\endgroup$ – Wesley Strik Jan 27 at 13:48
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    $\begingroup$ Yes, it's a form of a squeeze theorem $\endgroup$ – rtybase Jan 27 at 13:50
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    $\begingroup$ $e$xcellent answer. It feels more like a limit comparison result. $\endgroup$ – Wesley Strik Jan 27 at 13:50
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Use $\sqrt n + 1 \geq \sqrt{n+1}, n\in\mathbb N:$

$$a_n = a_{n-1}(1+\frac 1{\sqrt n})\implies a_n = \prod_{k=1}^n(1+\frac 1{\sqrt k})= \prod_{k=1}^n \frac{\sqrt k + 1}{\sqrt k} \geq \prod_{k=1}^n\frac{\sqrt{k+1}}{\sqrt k} = \sqrt{n+1}.$$

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  • $\begingroup$ That is a satisfying approach. $\endgroup$ – Wesley Strik Jan 27 at 16:09
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$a_n=\prod_{i=1}^n\left(1+\frac1{\sqrt{i}}\right).$ Now notice that $f(x)=1+x-e^{x/4}\ge 0$ for $x\leq 1.$ Hence $a_n\geq e^{\frac14\sum_{i=1}^n\frac1{\sqrt{i}}}\rightarrow \infty.$

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  • $\begingroup$ Clever, I would need some more practice to come up with something like this. $\endgroup$ – Wesley Strik Jan 27 at 13:52
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Another approach: $$a_n=a_{n-1}+\frac{1}{\sqrt{n}}a_{n-1}\geq a_{n-1}+\frac{1}{\sqrt{n}},$$ from which we quickly get $$a_n\geq\sum_{k=1}^n\frac{1}{\sqrt{k}}\geq\sum_{k=1}^n\frac{1}{k}$$ which is divergent, since the last sum gives the harmonic series.

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