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I am obliged to count the

$$\lim_{n\rightarrow\infty} a_n$$

where

$$a_n=\frac{1}{n^2+1}+\frac{2}{n^2+2}+\frac{3}{n^2+3}+\cdots+\frac{n}{n^2+n}.$$

I know that this type of task, I should use squeeze theorem. I should find two sequences one that is smaller and one that is bigger. So I find $\large b_n=\frac{1}{n^2+1}$, $ b_n \le a_n$, but I have trouble in finding the bigger sequence $c_n \ge a_n$. I will be glad for any help.

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4 Answers 4

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We will use the following: $$\sum_{k=1}^n k=\frac{n(n+1)}{2}.$$ Since $n\geq 1$, we have $$a_{n}=\frac{1}{n^2+1}+\frac{2}{n^2+2}+\frac{3}{n^2+3}+\cdots+\frac{n}{n^2+n}\geq\frac{1}{n^2+n}+\frac{2}{n^2+n}+\frac{3}{n^2+n}+\cdots+\frac{n}{n^2+n}=\frac{n(n+1)}{2(n^2+n)}$$ and $$a_{n}=\frac{1}{n^2+1}+\frac{2}{n^2+2}+\frac{3}{n^2+3}+\cdots+\frac{n}{n^2+n}\leq\frac{1}{n^2}+\frac{2}{n^2}+\frac{3}{n^2}+\cdots+\frac{n}{n^2}=\frac{n(n+1)}{2n^2}.$$ By the squeeze theorem, $\lim_{n\rightarrow\infty} a_n=\frac{1}{2}$.

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Hint: use, for $1\leq k\leq n$ : $$\frac{k}{n^2+n}\leq \frac{k}{n^2+k}\leq \frac{k}{n^2}.$$ The idea is to remove $k$ from the denominator.

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    $\begingroup$ @Marcin Majewski And keep in mind what $1+2+\cdots+n$ sums to. $\endgroup$ Jan 28, 2014 at 22:14
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Hint : $a_n= n^2(H_{\frac{n}{n+1}}+1$) where $H$ is a harmonic series, which in this case for large $n$ is $\sim -\frac{1}{n+1}+\frac{1}{2(n+1)^2}$.

Can you handle from here?

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Each term in the expression for $a_n$ is less than or equal to $\displaystyle \frac{n}{n^2+n} = \frac{1}{n+1}$, so I think the upper bound would be $$a_n = \frac{1}{n^2+1} + \frac{2}{n^2+2} + \cdots + \frac{n}{n^2+n} \le \frac{n}{n^2+n} + \frac{n}{n^2+n} + \cdots + \frac{n}{n^2+n} = \frac{n}{n+1} = c_n.$$

Of course, as $n \rightarrow \infty$, $c_n \rightarrow 1$.

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    $\begingroup$ To whoever downvoted, would you mind explaining WHY you did so? I would like to improve this response if possible. $\endgroup$
    – 2012ssohn
    Jan 28, 2014 at 22:07
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    $\begingroup$ @MarcinMajewski. This is not an answer. There are much better $b_n$ and $c_n$ that are really squeezing. Check the full and correct answer of alans, based on the same idea as mine. $\endgroup$
    – Tom-Tom
    Jan 28, 2014 at 22:22
  • $\begingroup$ What about a lower bound? You can't find a limit by squeezing without two bounds both approaching the limit. $\endgroup$ Jan 28, 2014 at 22:22

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