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

I am trying to solve the following inequality:


but I do not know how to do that. Please help me solve this inequality.

share|cite|improve this question
Should your first term be $2$? – Daniel Littlewood Nov 16 '12 at 18:46
@DanielLittlewood, It was correct. But I added 1 to both sides of inequality to be more clear. – Random Nov 16 '12 at 18:50
I've removed algebra tag, since we don't use algebra tag anymore, see meta for details. – Martin Sleziak Nov 18 '12 at 20:32
up vote 3 down vote accepted

Use the fact that $(1 + \frac1n)^{\frac1n} < 1 + \frac{1}{n^2}$ for $n > 1$. Thus your sum is $\sum_{k=1}^n (1 + \frac1k)^{\frac1k} < \sum_{k=1}^n (1 + \frac{1}{k^2}) < n + \frac{\pi^2}{6} < n+2$.

share|cite|improve this answer
$(1+1/n)^{1/n}>1+1/n^2$ not $(1+1/n)^{1/n}>1+1/n^2$. Now I think your proof will not work. – Amr Nov 16 '12 at 21:56
No, it's true that $(1+1/n)^{1/n} < 1 + 1/n^2$ for $n \ge 1$. To see this, just raise both sides to $n$th power, you get $1+1/n$ on the LHS and $(1+1/n^2)^n = 1+1/n + \cdots$ on the RHS. – Alan Guo Nov 19 '13 at 16:10

Since $\log(1+x)\leq x$ and $e^x\leq\frac{1}{1-x}$ we have: $$(1+1/k)^{1/k}=\exp\left(\frac{1}{k}\,\log(1+1/k)\right)\leq e^{1/k^2}\leq 1+\frac{1}{k^2-1}=1+\frac{1}{2}\left(\frac{1}{k-1}-\frac{1}{k+1}\right),$$ so: $$\sum_{k=2}^{n}(1+1/k)^{1/k} \leq (n-1)+\sum_{k=2}^n \frac{1}{k^2-1}\leq (n-1)+\sum_{k=2}^{+\infty}\frac{1}{k^2-1}=(n-1)+\frac{3}{4}< n.$$

share|cite|improve this answer

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.