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Let $a_{n} = \frac{1}{n+1} \left(1+ \cdots + \frac{1}{n} \right)$ for all natural number $n$. I don't know how to show $a_{n}$ converges to $0$. So far, I've figured out that the sequence is a dereasing sequence.

\begin{align*}\frac{a_{n+1}}{a_{n}} = \frac{n+1}{n+2} \frac{\left(1+ \cdots + \frac{1}{n} + \frac{1}{n+1} \right)}{\left(1+ \cdots + \frac{1}{n} \right)} &= \frac{n+1}{n+2} \left(1+\frac{\frac{1}{n+1}}{1+ \cdots + \frac{1}{n}} \right) \\ & \le \frac{n+1}{n+2} \left(1+ \frac{1}{n+1} \right) =1.\end{align*}

But, I'm not sure how to proceed with the solution from here. Any small hints or other solutions would be appreciated.

Gary
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gnahop
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1 Answers1

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In the second-to-last step you use the fact that $1 + \cdots + \frac{1}{n} \geq 1,$ but this is a very, very rough approximation. Being just a bit more careful here gives you what you want: for $n \geq 2$ we have $1 + \cdots + \frac{1}{n} \geq \frac{3}{2}$, so your second-to-last step gives $$\frac{a_{n+1}}{a_{n}} \leq \frac{n+1}{n+2}\left(1 + \frac{2}{3} \cdot \frac{1}{n+1} \right).$$ This should give you what you want.

ckefa
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