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I managed to prove that when $a_n \longrightarrow L$ then $\frac{n}{\frac{1}{a_1}+...+\frac{1}{a_n}} \longrightarrow L$, and $\frac{a_1+...+a_n}{n}\longrightarrow L$, and all these things scream the arithmetic/geometric mean inequality at me, but the only direction I could think of was getting an a larger sequence from the inequality to use as the upper bounding sequence in the squeeze theorem.

How can I get a lower bounding sequence to squeeze in the original sequence? Is there an alternative method?

daedsidog
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  • what if $a_1=0$? – Michael Apr 18 '19 at 15:58
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    You need the sequence to have $a_i>0$ for all $i$ to make sense. Then take $b_n=\log a_n$ and apply $b_n\to \log L.$ Apply that $\frac{b_1+\cdots+b_n}{n}\to\log L.$ You'll need some special case argument for when $L=0.$ – Thomas Andrews Apr 18 '19 at 15:58
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    $n/(1/a_1+...+1/a_n)\le\sqrt[n]{a_1...a_n}\le(a_1+...+a_n)/n$ when the $a_i>0$. – Thorgott Apr 18 '19 at 16:01
  • @Thorgott How do you know that the leftmost operand is smaller than the one in the middle? – daedsidog Apr 18 '19 at 16:22
  • Well, if you know the right inequality (which I believe you do), then apply it to the numbers $1/a_1,...,1/a_n$ (still positive) and rearrange to get the left inequality. – Thorgott Apr 18 '19 at 17:07

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