Limit of a sum

For positive integer $n$, define
$$f(n) = n + \frac{16 + 5n - 3n^2}{4n + 3n^2} + \frac{32 + n - 3n^2}{8n + 3n^2} + \frac{48 - 3n - 3n^2}{12n + 3n^2} + \cdots + \frac{25n - 7n^2}{7n^2}.$$
Then, the value of $\displaystyle\lim_{n\to\infty} f(n)$ is equal to

$$\lim_{n\to\infty}\left(\frac{n^2}{\left(n^2+1\right)(n+1)}+\frac{n^2}{\left(n^2+4\right)(n+2)}+\frac{n^2}{\left(n^2+9\right)(n+3)}+\dots+\frac{n^2}{\left(n^2+n^2\right)(n+n)}\right)$$
is equal to

$\displaystyle\lim_{n\to\infty}\sum_{r=1}^n\frac r{2r^2-7rn+6n^2}$ is equal to :

If $\displaystyle\lim_{n\to\infty}\frac{(n+1)^{k-1}}{n^{k+1}}\left[(nk+1)+(nk+2)+\dots+(nk+n)\right]=33\cdot\lim_{n\to\infty}\frac1{n^{k+1}}\cdot\left[1^k+2^k+3^k+\dots+n^k\right]$, then the integral value of k is equal to ______.

$\displaystyle\lim_{n\to\infty}\frac1{2^n}\left(\frac1{\sqrt{1-\frac1{2^n}}}+\frac1{\sqrt{1-\frac2{2^n}}}+\frac1{\sqrt{1-\frac3{2^n}}}+\dots+\frac1{\sqrt{1-\frac{2^n-1}{2^n}}}\right)$ is equal to

If $a=\displaystyle\lim_{n\to\infty}\sum_{k=1}^n\frac{2n}{n^2+k^2}$ and $f(x)=\sqrt{\dfrac{1-\cos x}{1+\cos x}}$, $x\in(0,1)$, then :

$\displaystyle\lim_{n\to\infty}\sum_{k=1}^n\frac{n(n^2+k^2)}{(n^4+n^2k^2+k^4)}$ is equal to

For $a\in\mathbb R$, $|a| > 1$, let

$$\lim_{n\to\infty}\left(\frac{1 + \sqrt[3]2 + \cdots + \sqrt[3]n}{n^{7/3}\left(\frac1{(an+1)^2} + \frac1{(an+2)^2} + \cdots + \frac1{(an+n)^2}\right)}\right) = 54.$$

Then the possible value(s) of $a$ is/are

For each positive integer $n$, let

$$y_n = \frac{1}{n}\left((n+1)(n+2)\cdots(n+n)\right)^{\frac{1}{n}}.$$

For $x \in \mathbb{R}$, let $[x]$ be the greatest integer less than or equal to $x$. If $\displaystyle\lim_{n \to \infty} y_n = L$, then the value of $[L]$ is _____ .

Let $f(x)=\displaystyle\lim_{n\to\infty}\left(\frac{n^n(x+n)\left(x+\frac n2\right)\cdots\left(x+\frac nn\right)}{n!\,(x^2+n^2)\left(x^2+\frac{n^2}{4}\right)\cdots\left(x^2+\frac{n^2}{n^2}\right)}\right)^{x/n}$, for all $x>0$. Then

For $a\in\mathbb R$ (the set of all real numbers), $a\ne-1$,
$$\lim_{n\to\infty}\frac{1^a+2^a+\cdots+n^a}{(n+1)^{a-1}\left[(na+1)+(na+2)+\cdots+(na+n)\right]}=\frac1{60}.$$
Then $a=$

Let
$$S_n=\sum_{k=1}^{n}\frac{n}{n^2+kn+k^2}\quad\text{and}\quad T_n=\sum_{k=0}^{n-1}\frac{n}{n^2+kn+k^2},$$
for $n=1,2,3,\ldots$. Then,