Compute $ \lim_{x \to\frac{\pi} {2}} \{1^{\sec^2 x} + 2^{\sec^2 x} + \cdots + n^{\sec^2 x}\}^{\cos^2 x} $ Find $\displaystyle \lim_{x \to\frac{\pi} {2}} \{1^{\sec^2 x} + 2^{\sec^2 x} + \cdots + n^{\sec^2 x}\}^{\cos^2 x} $
I tried to do like this :
Let $A=\displaystyle \lim_{x \to\frac{\pi}{2}} \{1^{\sec^2 x} + 2^{\sec^2 x} + \cdots + n^{\sec^2 x}\}^{\cos^2 x} $
Then $\ln A=\lim_{x \to\frac{\pi} {2}} \cos^2 x \ln\{1^{\sec^2 x} + 2^{\sec^2 x} + \cdots + n^{\sec^2 x}\}$
which implies
$$\ln A = \lim_{x \to\frac {\pi}{2}} \frac{1^{\sec^2 x} + 2^{\sec^2 x}+ \cdots + n^{\sec^2 x}}{\sec^2 x}$$
I'm stuck here. If I am on the right way please guide me to reach conclusion. Otherwise please describe the actual way. Thanks in advance.
 A: Let $f_n(x)$ be the sequence given by
$$f_n(x)=\left(\sum_{k=1}^n k^{\sec^2(x)}\right)^{\cos^2(x)}$$
Then, we have
$$\begin{align}
f_n(x)&=\left(n^{\sec^2(x)}\sum_{k=1}^n \left(\frac kn\right)^{\sec^2(x)}\right)^{\cos^2(x)}\\\\
&=n\left(\sum_{k=1}^n \left(\frac kn\right)^{\sec^2(x)}\right)^{\cos^2(x)}\\\\
\end{align}$$
For any fixed $n\ge1$
$$\lim_{x\to\pi/2}(k/n)^{\sec^2(x)}=\begin{cases}0&1\le k<n\\\\1&,k=n\end{cases}$$
Therefore, we have
$$\lim_{x\to \pi/2}f_n(x)=n$$
A: Since
$\sec(x) = 1/\cos(x)$
and
$\cos(\pi/2+y)
=\cos(\pi/2)\cos(y)-\sin(\pi/2)\sin(y)
=-\sin(y)
$,
$\begin{array}\\
a(n)
&=\lim_{x \to \pi/2}\left(\sum_{k=1}^n k^{\sec^2(x)}\right)^{\cos^2(x)}\\
&=\lim_{y \to 0}\left(\sum_{k=1}^n k^{\sec^2(y+\pi/2)}\right)^{\cos^2(y+\pi/2)}\\
&=\lim_{y \to 0}\left(\sum_{k=1}^n k^{1/\sin^2(y)}\right)^{\sin^2(y)}\\
&=\lim_{z \to 0}\left(\sum_{k=1}^n k^{1/z}\right)^{z}
\qquad z = \sin^2(y)\\
&=\lim_{w \to \infty}\left(\sum_{k=1}^n k^{w}\right)^{1/w}
\qquad w = 1/z\\
&=\lim_{w \to \infty}\left(n^w\sum_{k=1}^n (k/n)^{w}\right)^{1/w}\\
&=\lim_{w \to \infty}n\left(\sum_{k=1}^n (k/n)^{w}\right)^{1/w}\\
&=\lim_{w \to \infty}n\left(\sum_{k=0}^{n-1} ((n-k)/n)^{w}\right)^{1/w}\\
&=\lim_{w \to \infty}n\left(\sum_{k=0}^{n-1} (1-k/n)^{w}\right)^{1/w}\\
&=\lim_{w \to \infty}n\left(1+\sum_{k=1}^{n-1} (1-k/n)^{w}\right)^{1/w}\\
&\ge n\\
\text{and}\\
a(n)
&=\lim_{w \to \infty}n\left(1+\sum_{k=1}^{n-1} (1-k/n)^{w}\right)^{1/w}\\
&\le\lim_{w \to \infty}n\left(1+n-1\right)^{1/w}\\
&=\lim_{w \to \infty}nn^{1/w}\\
&\to n\\
\end{array}
$
so the limit is $n$.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
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 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[15px,#ffd]{\lim_{x \to \pi/2}\bracks{\large1^{\sec^{2}\pars{x}} + 2^{\sec^{2}\pars{x}} +
\cdots + n^{\sec^{2}\pars{x}}}^{\large\cos^{2}\pars{x}}}
\\[5mm] = &\
\lim_{x \to \infty}\pars{\sum_{k = 1}^{n}k^{x}}^{1/x} =
\exp\pars{\lim_{x \to \infty}{\ln\pars{\sum_{k = 1}^{n}k^{x}} \over x}}
\\[5mm] = &\
\exp\pars{\lim_{x \to \infty}{\sum_{k = 1}^{n}k^{x}\ln\pars{k} \over
\sum_{k = 1}^{n}k^{x}}}
\\[5mm] = &\
\exp\pars{\lim_{x \to \infty}
{\sum_{k = 1}^{n - 1}\pars{k/n}^{x}\ln\pars{k} + \ln\pars{n} \over
\sum_{k = 1}^{n - 1}\pars{k/n}^{x} + 1}} =
\expo{\ln\pars{n}} =
\bbox[15px,#ffd,border:1px solid navy]{n}
\end{align}
