How can I check if the sequence $x_{n}=\frac{n}{n^{2}+1}+\frac{n}{n^2+2}+...+\frac{n}{n^2+n},n=1,2,...$ converges?Is there a theorem that I could use?
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We have $1>x_n>\frac{n^2}{n^2+n}$ Now apply the squeeze theorem.
Ishan Banerjee
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Have you found this,using this relation: $$\frac{n}{n^2+n}+...+\frac{n}{n^2+1}<x_{n}<\frac{n}{n^2+1}+...+\frac{n}{n^2+1} \Rightarrow \frac{n^2}{n^2+n}<x_{n}<\frac{n^2}{n^2+1}$$ – evinda Feb 09 '14 at 14:43
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1Yes, I used almost exactly the same relation. – Ishan Banerjee Feb 09 '14 at 14:46
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Great!!!Thanks a lot for your answer!!! – evinda Feb 09 '14 at 14:48
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Notice the following: Every two integers $n,k$ such that $1\leq k \leq n$ satisfy $$\frac {1}{n+1} = \frac{n}{n^2+n} \leq \frac{n}{n^2+k} \leq \frac{n}{n^2}=\frac{1}{n}$$ So we get $$ \frac{1}{1+\frac{1}{n}} = n \cdot \frac {1}{n+1} \leq x_n \leq n \cdot \frac{1}{n} =1$$ Thus by the Sandwich Theorem we get $x_n\rightarrow 1$.
DalyaG
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$$\frac{n}{n^2}+\ldots+\frac{n}{n^2+n}<x_n<\frac{n}{n^2}+\ldots+\frac{n}{n^2}\implies \lim_{n\to \infty}\frac{n}{n}<\lim_{n\to\infty}x_n<\lim_{n\to\infty}\frac{n}{n+1}$$
We can re-write it as,
$$1<\lim_{n\to \infty}x_n<1$$
So, $x_n=1$
Hawk
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I understand..Could I also take this relation: $\frac{n}{n^2+n}+...+\frac{n}{n^2+1}<x_{n}<\frac{n}{n^2+1}+...+\frac{n}{n^2+1} \Rightarrow \frac{n^2}{n^2+n}<x_{n}<\frac{n^2}{n^2+1}$ ? – evinda Feb 09 '14 at 14:40
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1@evinda Certainly you could use that, the point is to get an inequality from which we can have the left inequality to be equal to the right inequality. – Hawk Feb 09 '14 at 14:43
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@evinda I think you meant $\frac{n}{n^2+n}$ instead of $\frac{n}{n^2+1}$ on your left inequality. – Hawk Feb 09 '14 at 14:45
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Use Riemann Sums so the sequence should converge to an integral.
Waseem Francis
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Have you given any thought towards which integral this would be a Riemann sum for? – Erick Wong Feb 09 '14 at 18:23
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no, but I can see that by sandwich you can solve it. if you insist I can calculate it for you in the near time I just have no time now. you can start calculating it by getting out $\frac{1}{n}$ – Waseem Francis Feb 09 '14 at 19:00