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Here's an interesting sequence:

$a_{n}=\sum_{k=0}^n \frac{1}{n+k}$

And my task is to find whether it has a $\lim$ as $n\rightarrow\infty$ or not. My first strategy was analyzing this sequence. So you can prove that this sequence is decreasing (by discovering that $a_{n+1}=a_{n}-\frac{1}{n}+\frac{1}{2n+1}+\frac{1}{2n+2}$ and this term that you are subtracting from $a_{n}$ must be less than zero), and you can clearly see it's always positive (so it can't be divergent to $\infty$ nor $-\infty$). So intuitively, this has some $\lim$ but I can't prove it formally yet. (My task is not to find the exact $\lim$ but rather say whether it exists or not).

So then I tried to find the exact $\lim$ and my intuition told me to use Squeeze Theorem to prove it. But I'm not sure whether it's a good idea because after an hour I still haven't found anything that could help me progress further.

So, having proven that $a_{n}>0$ and $a_{n}$ is decreasing what should I do in order to prove that the limit exists (or show that it doesn't)?

qiubit
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2 Answers2

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$$\sum_{k=0}^n\frac1{n+k}=\frac1n\sum_{k=0}^n\frac1{1+\frac kn}\xrightarrow[n\to\infty]{}\int\limits_0^1\frac{dx}{1+x}$$

Another way:

$$\frac1n+\frac1{n+1}+\ldots+\frac1{2n}\le \frac1n+\frac1n+\ldots+\frac1n= \frac nn=1$$

and since the sequence is clearly monotone ascending it converges.

Timbuc
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  • I can't use integrals yet – qiubit Oct 30 '14 at 17:19
  • @JohnZHANG , you are right. Thank you. I was hesitating between using the function $;\frac1x;$ between $;1,2;$ or the function $;\frac1{1+x};$ between $;0,1;$ and mixed things. – Timbuc Oct 30 '14 at 17:26
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Hints:

A decreasing sequence with bounded blow is always convergent. (Also an increasing sequence with bounded sup is always convergent.

Paul
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