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Evaluate the following limit:


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

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up vote 8 down vote accepted

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. – alex.jordan Sep 8 '12 at 16:12
@alex.jordan: The hint is excellent, it leaves the OP something to do. – André Nicolas Sep 8 '12 at 16:14
Ah, my comment stemmed from not seeing how the hint could help :) +1 – alex.jordan Sep 8 '12 at 16:22
@alex.jordan The hints says, for all $n$, that $$ \sum_{r=1}^n \frac{r}{n^2+n+n} <leq \sum_{r=1}^n \frac{r}{n^2+n+r} <leq \sum_{r=1}^n \frac{r}{n^2+n}$$ Now consider what happens when $n$ grows large. What happens to the difference between the upper limit and lower limit? – Sasha Sep 8 '12 at 16:28
Best method i think +1 – mick Sep 8 '12 at 19:01

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. – André Nicolas Sep 8 '12 at 16:21

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