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I am to determine if the sequence

$$x_n = \frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n-1}+\frac{1}{2n} $$

converges or not?

I'm thinking I if I can determine if the sequence has some kind om bound and if it is either increasing or decreasing? Because then I know from the Monotone convergence theorem that it converges.

But I'm not sure how you would prove this.

Ahmad Bazzi
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F Wi
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3 Answers3

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Hint:

You may write $$\sum_{k=1}^n\frac{1}{n+k} = \frac{1}{n}\sum_{k=1}^n\frac{1}{1+\frac{k}{n}}\stackrel{n \to \infty}{\longrightarrow} \int_0^1\frac{1}{1+x}\; dx$$

trancelocation
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  • How do I get to $$\sum_{k=1}^n\frac{1}{n+k}$$ – F Wi Jan 29 '19 at 13:12
  • @FWi : It is not quite clear to me what you are asking. – trancelocation Jan 29 '19 at 13:15
  • I don't understand how the sequence can be written as a sum like that – F Wi Jan 29 '19 at 13:16
  • $$\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n-1}+\frac{1}{2n} =\frac{1}{n(1+\frac{1}{n})}+\frac{1}{n(1+\frac{2}{n})}+...+\frac{1}{n(1 + \frac{n}{n})} = \frac{1}{n}\left( \frac{1}{1+\frac{1}{n}} + \frac{1}{1+\frac{2}{n}} + \cdots + \frac{1}{1+\frac{n}{n}}\right)$$ – trancelocation Jan 29 '19 at 13:23
1

Correct if wrong :

$1/2=n/(2n) <x_n < n/(n+1) <1,$

hence bounded.

$x_n = 1/(n+1)+..........1/(2n);$

$x_{n+1}= 1/(n+2) +.....1/(2n+1)+1/(2n+2).$

Note : $1/(2n+2)= (1/2)(1/(n+1)),$ and

$1/(2n+1)>1/(2n+2)=$

$(1/2)(1/(n+1))$.

$1/(2n+1)+1/(2n+2) >$

$1/(n+1).$

Hence

$x_{n+1} > x_n.$

$x_n$ is bounded, increasing , hence convergent.

Peter Szilas
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The sequence is bounded, since

$$x_n = \frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n-1}+\frac{1}{2n} \geq \frac{1}{2n}+\frac{1}{2n} + \cdots + \frac{1}{2n} = \frac12$$

and

$$x_n = \frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n-1}+\frac{1}{2n} \leq \frac{1}{n+1}+\frac{1}{n+1} + \cdots + \frac{1}{n+1} = \frac{n}{n+1}\leq 1$$

To see whether it is monotonic, look at the sign of

$$x_n - x_{n-1}.$$

The expression $x_n-x_{n-1}$ should have a lot of things cancel out.

5xum
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  • This answer recieved a downvote. I would appreciate it if the person that downvoted my answer explains why they consider my answer to be incorrect. I will gladly correct any part of the answer that is incorrect. – 5xum Jan 29 '19 at 12:43
  • Why the negative vote? – Crostul Jan 29 '19 at 13:18
  • @Crostul It happens sometimes. One thing I notice is that I recieve a downvote on my answer usually soon after I give a downvote to some other answer. Not always, but the correlation is there, I'm almost sure of it. That offers one possible explanation for this downvote, since another user answered this same question, and I downvoted their answer (while explaining why it was incorrect). Now, of course, that's not the only explanation, but statistically, this explanation is right at least some of the time. – 5xum Jan 29 '19 at 13:21
  • Have you factored out some term to get the inequalities above? – F Wi Jan 29 '19 at 13:32
  • @FWi I didn't factor anything out. I used the fact that $\frac{1}{n+2}\leq \frac{1}{n+1}$, and the fact that $\frac{1}{n+3}\leq \frac{1}{n+1}$, and so on. – 5xum Jan 29 '19 at 13:34
  • this may seem like a very trivial question but how is $\frac{n}{n+1}\leq 1$ ? – F Wi Jan 29 '19 at 17:52
  • I think maybe you mean strictly less than 1? – F Wi Jan 29 '19 at 18:55
  • @FWi if $a<b$, then surely, $a\leq b$ is also true, isn't it? – 5xum Jan 29 '19 at 21:02