Indeterminate form?

Calculus Level 4

$\large \lim_{n\rightarrow \infty} \dfrac{1}{n} \displaystyle\sum_{r=1}^{2n} \dfrac{r}{\sqrt{n^2+r^2}} = \sqrt{a} - 1, \ \ \ \ \ \ \ a = \ ?$

The answer is 5.

1 solution

Aniket Verma
Mar 16, 2015

$\text{lim}_{n\rightarrow \infty} \dfrac{1}{n} \displaystyle\sum_{r=1}^{2n} \dfrac{r}{\sqrt{n^2+r^2}} = \sqrt{a} - 1$

$\implies$ $lim_{n\rightarrow \infty} \dfrac{1}{n} \displaystyle\sum_{r=1}^{2n} \dfrac{1}{\sqrt{\dfrac{n^2}{r^2}+1}} = \sqrt{a} - 1$

now the above equation has changed to a problem of limit as sum.

convert $lim_{n\rightarrow \infty} \displaystyle\sum_{r=1}^{2n} ~into~ \int_{0}^{2} , \dfrac{r}{n} ~into~ x~ and \dfrac{1}{n} ~into ~ dx$

$now,$ $\text{lim}_{n\rightarrow \infty} \dfrac{1}{n} \displaystyle\sum_{r=1}^{2n} \dfrac{r}{\sqrt{n^2+r^2}}$ $=$ $\displaystyle \int_{0}^{2} \dfrac{x}{\sqrt{x^2 + 1}} dx$ = $[\sqrt{x^2 + 1}]_{0}^{2} = \sqrt{5} - 1$

$therefore \rightarrow ~ a = 5$

I think that your solution lacks a bit .

You must mention that you have taken $\frac{r}{n}=x , \frac{1}{n}=dx$

Also there's a small error in your solution, it should be $\sqrt{x^{2}+1}$ , you have not written the square root symbol .

If you want some Latex help ,

\therefore yields $\therefore$

Try using \displaystyle to make your Latex look better

For example :

Without it $\int$

With displaystyle $\displaystyle \int$

I do hope that I was useful :)

A Former Brilliant Member - 6 years, 2 months ago

Thanks a lot for the suggestion.

Aniket Verma - 6 years, 2 months ago

You are welcome $\ddot\smile$

A Former Brilliant Member - 6 years, 2 months ago

Indeterminate form?

The answer is 5.

1 solution

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