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Evaluate the following limit:

$$\lim_{n\to\infty}\left(\sum_{r=1}^{n}{\frac{r}{n^{2}+n+r}}\right)$$

The answer given is $\frac{1}{2}$.

2 Answers 2

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Squeeze it: $$ \frac{r}{n^2+n+n}\leq\frac{r}{n^2+n+r}\leq\frac{r}{n^2+n} $$

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    OP wants the sum of the series, not the limit of the sequence.2012-09-08
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    @alex.jordan: The hint is excellent, it leaves the OP something to do.2012-09-08
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    Ah, my comment stemmed from not seeing how the hint could help :) +12012-09-08
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    @alex.jordan The hints says, for all $n$, that $$ \sum_{r=1}^n \frac{r}{n^2+n+n} Now consider what happens when $n$ grows large. What happens to the difference between the upper limit and lower limit?2012-09-08
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    Best method i think +12012-09-08
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    @Sasha Yes, My reply to Andre was meant to convey that I now understood clark's hint.2012-09-09
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Yes I have got it thank you. First put the expression as $\frac{r}{n^2+n+n}$ and then evaluate the limit. It will become $\frac{1}{2}$. Now put the given expression as $\frac{r}{n^2+n}$. Again the value will come out to be $\frac{1}{2}$. So from Sandwich Theorem we will have the limit as $\frac{1}{2}$. Thanks again

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    In a course, **much** more detail would need to be supplied, like an explicit summation.2012-09-08