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How can we determine the sum

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

I tried to reduce this to an integral problem by dividing both numerator and denominator by $n^2$ but we get the term $\dfrac{\frac{1}{n}}{1+\frac{r}{n^2}}$ where $r$ ranges from $1$ to $n$ but to get the variable $x$ in integration we would need $\frac{r^2}{n^2}=x^2$. So,i am not sure how to do this. I will be very grateful for the solutions.

Cheese Cake
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madness
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1 Answers1

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We have $$\frac{n^2}{n^2+n} \leq\left(\frac{n}{n^2+1}+\frac{n}{n^2+2}+\cdots+\frac{n}{n^2+n}\right) \leq \frac{n^2}{n^2+1},$$ so by the squeeze theorem $$\lim_{n \to \infty} \left(\frac{n}{n^2+1}+\frac{n}{n^2+2}+\cdots+\frac{n}{n^2+n}\right)=1.$$

i like math
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  • Thank you very much. A bit of explanation of how the inequality was derived will be very helpful for me. – madness Sep 11 '22 at 08:57
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    Sure! There are $n$ terms in the sum, so I found the lower bound by multiplying the smallest term by $n$ and the upper bound by multiplying the biggest term by $n$. – i like math Sep 11 '22 at 08:58