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Given the sequence $$\frac{1+2+...+n}{n+2}-\frac{n}{2}$$ I am asked to check if it converges. How can I do this? One way is to check if the sequence is bounded and monotonic, right? But how could we see if the sequence is monotonic?

Mary Star
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2 Answers2

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The numerator can be written as $\frac{n^2+n}{2}$ (that's the closed formula Daniel is hinting at) and so if you combine the fractions into one, you get a numerator of $-n$ and a denominator of $2(n+2)$ Taking the limit for $n$ to inf gives -0.5

imranfat
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  • With this way we find the limit without having shown that the sequence converges, right? To do it with the other way, to show that it's bounded and monotonic, is the following correct? (By writing it as you said, we have $\frac{-n}{2(n+2)}$ $$\frac{-n-1}{2(n+3)}-\frac{-n}{2(n+2)}=-(\frac{n+1}{2(n+3)}+\frac{n}{2(n+2)})=- \frac{(n+1)(n+2)+n(n+3)}{2(n+2)(n+3)} \leq 0$$ $\frac{(n+1)(n+2)+n(n+3)}{2(n+2)(n+3)}$ is positive and with the $-$ it's negative. So the sequence is decreasing. – Mary Star Feb 10 '14 at 21:58
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    Yes, and when you take the derivative of the resulting term, you can see that it is decreasing with -0.5 as the limit. You did do a typo, it's $n+2$ not $n+3$ – imranfat Feb 10 '14 at 22:07
  • But at the first term of the difference we take $n+1$, so it gets $2((n+1)+2)=2(n+3)$, or am I wrong? – Mary Star Feb 10 '14 at 22:12
  • @MaryStar after putting $\frac{n^2+n}{2}$ in the numerator, I made common denominators and combined the numerators. The common denominator is $2n+2$ and the new numerator turns out to be just $-n$ – imranfat Feb 11 '14 at 15:17
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If you name $u_n$ a term of the sequence, you can check the sign of $u_{n+1}-u_n$, if the sign is always negative (resp. positive), the sequence is decreasing (resp. increasing).

Gurvan
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  • But how can find this difference? $u_{n+1}-u_n=\frac{1+2+...+n+(n+1)}{n+3}-\frac{n+1}{2}-\frac{1+2+...+n}{n+2} + \frac{n}{2} $.How can I continue? – Mary Star Feb 10 '14 at 21:39
  • Look at the others answers and find another way to write $u_n$ (actually they gave it to you). – Gurvan Feb 10 '14 at 21:42