If $f(k)=\dfrac{(k+1)^{k+1}}{k^k}\sum_{t=k+1}^{\infty}\dfrac{1}{t^2}$ then $f(k+1)>f(k)$ 
Let $$f(k)=\dfrac{(k+1)^{k+1}}{k^k}\sum_{t=k+1}^{\infty}\dfrac{1}{t^2}.$$
  Prove $$f(k+1)>f(k).$$

My idea:
$$f(k+1)-f(k)=\dfrac{(k+2)^{k+2}}{(k+1)^{k+1}}\sum_{t=k+2}^{\infty}\dfrac{1}{t^2}-\dfrac{(k+1)^{k+1}}{k^k}\sum_{t=k+1}^{\infty}\dfrac{1}{t^2}$$
and following is very ugly.
I think someone can use nice methods. Thank you.
This problem is from this topic because
$$f(k)=\dfrac{(k+1)^{k+1}}{k^k}\sum_{t=k+1}^{\infty}\dfrac{1}{t^2}<\dfrac{(k+1)^{k+1}}{k^k}\left(\dfrac{1}{k}-\dfrac{1}{n}\right)$$
and note that
$$\lim_{k\to\infty}\dfrac{(k+1)^{k+1}}{k^k}\dfrac{1}{k}=e.$$
 A: First, recall the series appear in $f(k)$ has an integral representation:
$$\sum_{s=1}^{\infty}\frac{1}{(k+s)^2} = \int_{0}^{\infty}\frac{t e^{-kt}}{e^t-1}dt
=\int_{0}^{\infty}\frac{2t e^{-2kt}}{e^{2t}-1}d2t
=4\int_0^{\infty}\frac{t}{e^t-e^{-t}} e^{-(2k+1)t} dt
$$
Consider the integral $I(\eta)$ defined below and integrate it by parts:
$$\begin{align}
I(\eta) =& \int_{0}^{\infty}\frac{\eta t}{e^t - e^{-t}} e^{-\eta t}dt
 = \int_0^{\infty}\frac{t}{e^t - e^{-t}}d(1 - e^{-\eta t})\\
=& \left[\frac{t}{e^t - e^{-t}}(1 - e^{-\eta t})\right]_0^{\infty} - \int_0^{\infty}(1-e^{-\eta t}) d (\frac{t}{e^t - e^{-t}})\\
=&\int_0^{\infty}(1-e^{-\eta t}) \varphi(t) dt
\end{align}$$
where 
$\varphi(t) = (\frac{-t}{e^t - e^{-t}})' = \frac{t(e^t+e^{-t})}{(e^t-e^{-t})^2} - \frac{1}{e^t-e^{-t}}$. 
The integrand of $I(\eta)$ has two factors. If one fix $t$ and consider the first factor $1 - e^{-\eta t}$ as a function of $\eta$, it is strictly increasing. For the second factor $\varphi(t)$, it is easy to check $\varphi(t) > 0$ for all $t > 0$. Combine these, we can conclude 
$I(\eta)$ is a increasing function in $\eta$. 
As a consequence, we obtain:
$$(k + \frac12)\sum_{t=k+1}^{\infty}\frac{1}{t^2} = 2I(2k+1)
\le 2I(2k+3) = (k + \frac32)\sum_{t=k+2}^{\infty}\frac{1}{t^2}
\tag{*1}$$
Second, consider the function:
$$\psi(x) := (x+\frac12)\log(x+\frac12) - (x-\frac12)\log(x-\frac12)-\log x$$
We have:
$$\begin{align}
\psi'(x) &= \log(x+\frac12)-\log(x-\frac12)-\frac{1}{x} = \log\left(\frac{1+\frac{1}{2x}}{1-\frac{1}{2x}}\right) - \frac{1}{x}\\
\psi''(x) &= -\frac{1}{x^2(4x^2-1)} 
\end{align}$$
What is sort of obvious is $\psi'(x) \to 0$ as $x \to \infty$ and $\psi''(x) < 0$ for $x > \frac12$. From this we can conclude $\psi'(x) > 0$ for $x \in (\frac12,\infty)$ and $\psi(x)$ is an increasing function there. As a consequence, we get:
$$\frac{(k+1)^{k+1}}{k^k (k+\frac12)} = e^{\psi(k+\frac12)} \le e^{\psi(k+\frac32)}
  = \frac{(k+2)^{k+2}}{(k+1)^{k+1}(k+\frac32)}\tag{*2}$$
Multiply $(*1)$ with $(*2)$ gives us $f(k) \le f(k+1)$ immediately.
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