How prove $\frac{(k+1)^{k+1}}{k^k}\sum_{t=k+1}^{n}\frac{1}{t^2}<e$

Let $k,n\in \mathbb{N},n\ge k$, prove that $$\dfrac{(k+1)^{k+1}}{k^k}\sum_{t=k+1}^{n}\dfrac{1}{t^2}<e.$$

I got the impression that this inequality is very sharp.

My idea: $$\sum_{t=k+1}^{n}\dfrac{1}{t^2}\le\sum_{t=k+1}^{n}\dfrac{1}{t(t-1)}=\dfrac{1}{k}-\dfrac{1}{n},$$

$$\Longleftrightarrow \dfrac{(k+1)^{k+1}}{k^k}\left(\dfrac{1}{k}-\dfrac{1}{n}\right)<\dfrac{(k+1)^{k+1}}{k^k}\dfrac{1}{k}<e$$ $$\Longleftrightarrow \left(1+\dfrac{1}{k}\right)^{k+1}<e.$$ It is well konwn that $$(1+1/x)^{x+1}>e,x>0.$$ so I failed at this direction. Thank you everyone help.

Idea 2: use the well-known $$\left(1+\dfrac{1}{x}\right)^x<e\left(1-\dfrac{1}{2(1+x)}\right)$$ $$\Longleftrightarrow \dfrac{n-k}{k+1}\left(1-\dfrac{1}{2(k+1)}\right)<1,$$ but this can not prove the inequality either.

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math110: I rephrased some of the English, please see if you are okay with it. –  Shuhao Cao Jul 4 '13 at 2:59
Oh,Thank you,@ShuhaoCao –  math110 Jul 4 '13 at 3:05
The Maple command $$asympt((k+1)^{k+1}sum(1/j^2, j = k+1 .. infinity))/k^k, k, 4)assuming \, k::posint$$ produces $${{\rm e}^{1}}-1/8\,{\frac {{{\rm e}^{1}}}{{k}^{2}}}+O \left( {k}^{-3} \right)$$ This implies the inequality under consideration for big $k$. –  user64494 Jul 4 '13 at 13:45

$$\sum_{r=m+1}^{\infty} \frac{1}{r^2}=\int_0^1 \frac{x^m \ln x}{x-1}\,dx$$
\begin{aligned}\left(1+\frac{1}{k}\right)^{k+1} \sum_{t=k+1}^n \frac{k}{t^2}&<\left(1+\frac{1}{k}\right)^{k+1} \sum_{t=k+1}^{\infty} \frac{k}{t^2}\\&=\left(1+\frac{1}{k}\right)^{k+1}\int_0^1 \frac{kx^k \ln x}{x-1}\,dx\\&<\left(1+\frac{1}{k}\right)^{k+1}\int_0^1 kx^{k-\frac{1}{2}} \,dx\\&=\left(1+\frac{1}{k}\right)^{k+1}\left(\frac{2k}{2k+1}\right)\\&=\left(1+\frac{1}{k}\right)^k\frac{2k+2}{2k+1}\\&<e\end{aligned}
Now,I have other idea: if we pf:$$f(k+1)>f(k)$$,then this inequality have prove it,where $$f(k)=\dfrac{(k+1)^{k+1}}{k^k}\sum_{t=k+1}^{\infty}\dfrac{1}{t^2}$$ –  math110 Jul 4 '13 at 3:58
and my question how prove $$f(k+1)\ge f(k)?$$ –  math110 Jul 4 '13 at 3:59