Evaluate Limit: $\lim_{n\rightarrow \infty } n(\frac{1}{(2n+1)^2} + \cdots)$

[lfdahl]

Evaluate the limit:

$\lim_{n\rightarrow \infty }\left ( n\left (\frac{1}{\left ( 2n+1 \right )^2}+\frac{1}{\left ( 2n+3 \right )^2}+ ...+\frac{1}{\left ( 4n-1 \right )^2} \right ) \right )$

Hint:

Spoiler

Consider the function: $f(x)=\frac{1}{x^2}$ on a suitable interval ...

 

[Saitama]

Evaluate the limit:

$\lim_{n\rightarrow \infty }\left ( n\left (\frac{1}{\left ( 2n+1 \right )^2}+\frac{1}{\left ( 2n+3 \right )^2}+ ...+\frac{1}{\left ( 4n-1 \right )^2} \right ) \right )$

Hint:

Spoiler

Consider the function: $f(x)=\frac{1}{x^2}$ on a suitable interval ...$

Spoiler

Rewrite the expression as:
$$\lim_{n\rightarrow \infty} \frac{1}{n}\left(\frac{1}{(2+\frac{1}{n})^2}+\frac{1}{(2+\frac{3}{n})^2}+\frac{1}{(2+\frac{5}{n})^2}+\cdots +\frac{1}{(2+\frac{2n-1}{n})^2}\right)$$
$$=\lim_{n\rightarrow \infty} \frac{1}{n}\sum_{r=1}^{n} \frac{1}{(2+\frac{2r-1}{n})^2}$$
$$=\lim_{n\rightarrow \infty} \frac{1}{n}\sum_{r=1}^{n} \frac{1}{(2+\frac{2r}{n}-\frac{1}{n})^2}$$

Since $n\rightarrow \infty$, $1/n \rightarrow 0$ and the above limit can be written as the following definite integral:
$$\int_0^1 \frac{dx}{(2+2x)^2}=\frac{1}{4}\left(-\frac{1}{1+x}\right|_0^1=\boxed{\dfrac{1}{8}}$$

 

[lfdahl]

Spoiler

Rewrite the expression as:
$$\lim_{n\rightarrow \infty} \frac{1}{n}\left(\frac{1}{(2+\frac{1}{n})^2}+\frac{1}{(2+\frac{2}{n})^2}+\frac{1}{(2+\frac{3}{n})^2}+\cdots +\frac{1}{(2+\frac{2n-1}{n})^2}\right)$$
$$=\lim_{n\rightarrow \infty} \frac{1}{n}\sum_{r=1}^{2n-1} \frac{1}{(2+\frac{r}{n})^2}$$
The above limit can be written as the following definite integral:
$$\int_0^2 \frac{dx}{(2+x)^2}=\left(-\frac{1}{2+x}\right|_0^2=\frac{-1}{4}+\frac{1}{2}=\boxed{\dfrac{1}{4}}$$

Good (and fast!) job, Pranav! - but:

Spoiler

You´re missing a factor 2 somewhere ...

 

[Saitama]

Good (and fast!) job, Pranav! - but:

Spoiler

You´re missing a factor 2 somewhere ...

Sorry about that, I misread the problem.

Spoiler

Do you mean a factor of 1/2?

 

[lfdahl]

Yes

 

[lfdahl]

Solution by other:

Spoiler

Consider the function $f(x) = \frac{1}{x^2}$ on the interval [2;4].
If we divide this interval into $n$ subintervals of equal length $\frac{2}{n}$ using the points $x_k= 2 + \frac{2k}{n}$, $0 \le k \le n$, and choose our sample points to be the midpoints $c_k= 2 + \frac{2k-1}{n}$,
$1\le k \le n$, of these subintervals, the Riemann sum for these data becomes
\[\sum_{k=1}^{n}f(c_k)\Delta x_k = \sum_{k=1}^{n}f\left ( 2+\frac{2k-1}{n} \right )\frac{2}{n}=\sum_{k=1}^{n}\frac{2n}{(2n+2k-1)^2}\]
And the definition of definite integral gives
\[\lim_{n\rightarrow \infty }\sum_{k=1}^{n}\frac{2n}{(2n+2k-1)^2} =\int_{2}^{4}\frac{dx}{x^2}= \left [ -\frac{1}{x} \right ]_{2}^{4}=\frac{1}{4}\]
Therefore
\[\lim_{n\rightarrow \infty }\left ( n\left (\frac{1}{(2n+1)^2}+\frac{1}{(2n+3)^2}+ ...+\frac{1}{(4n-1)^2} \right ) \right )=\frac{1}{8}\]