Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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

A sketch. First note that:

$$\sum_{r=m+1}^{\infty} \frac{1}{r^2}=\int_0^1 \frac{x^m \ln x}{x-1}\,dx$$


$$\begin{aligned}\left(1+\frac{1}{k}\right)^{k+1} \sum_{t=k+1}^n \frac{k}{t^2}&<\left(1+\frac{1}{k}\right)^{k+1} \sum_{t=k+1}^{\infty} \frac{k}{t^2}\\&=\left(1+\frac{1}{k}\right)^{k+1}\int_0^1 \frac{kx^k \ln x}{x-1}\,dx\\&<\left(1+\frac{1}{k}\right)^{k+1}\int_0^1 kx^{k-\frac{1}{2}} \,dx\\&=\left(1+\frac{1}{k}\right)^{k+1}\left(\frac{2k}{2k+1}\right)\\&=\left(1+\frac{1}{k}\right)^k\frac{2k+2}{2k+1}\\&<e\end{aligned}$$

share|cite|improve this answer
Nice!It's very nice!+1 – math110 Jul 4 '13 at 3:45
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.