Prove that $(1+\frac{1}{n})^k &lt; 1+ \frac{k}{n}+\frac{k^2}{n^2}, \forall$ n, k nonnegative integers, $k\le n$. I know that has something to do with Bernoulli's inequality $(1+\alpha)^x\ge 1+\alpha x, \alpha \ge -1, n\ge1$. If I reconsider Bernoulli's inequality with $\alpha=\frac{1}{n}$ and $x=k$ it follows that $(1+\frac{1}{n})^k\ge 1+\frac{k}{n}$, but I don't know how to continue. I also tried to prove it with induction where I consider $p(n):(1+\frac{1}{n})^k &lt; 1+ \frac{k}{n}+\frac{k^2}{n^2}$ to be true and prove that $ p(n+1):(1+\frac{1}{n+1})^k &lt; 1+ \frac{k}{n+1}+\frac{k^2}{(n+1)^2}$ to be also true, but it didn't work. Thank you!
123856
We know that $\dfrac1n&gt;0$, $\dfrac1n&lt;\dfrac mn&lt;1$, $\left(\dfrac1n+1\right)^m &lt; \left(\dfrac mn+1\right)^m$, so we have to prove that $\left(\dfrac mn+1\right)^m&lt;\dfrac mn+1+\left(\dfrac mn\right)^2$.  I tried using Newton's binomial expression of $\left(\dfrac mn+1\right)^m$ and calculating the difference, but I didn't get there.
47306
How do I prove the following inequality :  $$\Bigg(\frac{2}{\alpha^2} \, \big( e^{\alpha x} - e^{\alpha y} \big) \, + \, e^{\alpha y} (y^2 - x^2) \; \Bigg) &gt; 0 $$  given, $x, y &gt; 0$ ?  Can anyone provide me with hints about this problem ?  Thanks in advance.  Here, $\alpha$ is a positive constant.
111670
I have got an inequality problem which is as follow:     Show that $e^n&gt;\frac{(n+1)^n}{n!}$   I can do it by induction but I have been told to prove it without induction.  My Work:  $$e^n=1+n+\frac{n^2}{2!}+\frac{n^3}{3!}+........$$ $$e^n&gt;1+n+\frac{n^2}{2!}+\frac{n^3}{3!}+........+\frac{n^n}{n!}$$ $$e^n&gt;\frac{n^n}{n!}+\frac{n^{n-1}}{(n-1)!}.......+\frac{n^2}{2!}+n+1$$  From here I can't go further.   I shall be thankful if you guys can provide me a complete solution/proof of this inequality. A hint will also work.  Thanks in advance.
47364
Note: I don't want to get the full solution, but only a hint.  I have to show that for $x \in [0, \infty)$, the sequence $\left(1+\frac xk\right)^k$ is monotonically increasing. We were already given the hint that we could use the inequality of arithmetic and geometric means, but I don't see how to apply it yet.   I tried to do it by induction. While the case $k = 1$ works easily, I don't know how to start further from here. I already tried to use the definition of the binomial theorem, but it didn't lead me anywhere.
17401
So I'm a bit stuck on the following problem I'm attempting to solve. Essentially, I'm required to prove that $\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2} &lt; 1$ for all $n$. I've been toiling with some algebraic gymnastics for a while now, but I can't seem to get the proof right. Proving it using calculus isn't a problem, but I'm struggling hither.
86737
So I have to prove $$ \frac{a+b+c}{abc} \leq \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}.$$  I rearranged it   $$ a^2bc + ab^2c + abc^2 \leq b^2c^2 + a^2c^2 + a^2b^2 .$$  My idea from there is somehow using the AM-GM inequality. Not sure how though. Any ideas?  Thanks
471
How to prove for $s&lt;1$$$|a+b|^s\le|a|^s+|b|^s$$ I tried to prove $|a+b|\le (|a|^s+|b|^s)^{\frac{1}{s}}$, but $\frac{1}{s}$ may not be integer, so I do not know how to expand it if it is not integer.  Is it the right way to prove it by trying to expand it using binomial expansion? How to prove it?  Thanks!
13042
Trying to prove that above stated question for $n \geq 1$. A hint given is that you should use $\frac{1}{(2k-1)(2k+1)}=\frac{1}{2}(\frac{1}{2k-1}-\frac{1}{2k+1})$. Using this, I think I reduced it to $\frac{1}{2}(\frac{1}{n^2}-(\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\cdots+\frac{1}{2n+1}))$. Just not sure if it's correct, and what to do with the second half.
70751
Let $a$, $b$ and $c$ be positive real numbers with $abc=1$. Prove that   $$ \frac{a^{n+2}}{a^n+(n-1)b^n}+\frac{b^{n+2}}{b^n+(n-1)c^n}+\frac{c^{n+2}}{c^n+(n-1)a^n} \geq \frac{3}{n} $$   for each integer $n$.   I have used Cauchy-Schwarz inequalities and Jensen inequality. But I am stuck. I need some idea and advice on this problem. Induction would be cruel.
124186
$$\sum_{k=n}^\infty{\frac{1}{k!}} \leq \frac{2}{n!}$$  Can someone show why this estimate holds true? I tried quite a bit but couldn't really find a way to approach this.   says it is true but I don't know what the gamma function is.  $$ \sum_{k=n}^\infty{\frac{1}{k!}} = \frac{1}{n!} + \sum_{k = n+1}^\infty \frac{1}{k!}$$ So then I need to show that$$ \sum_{k=n+1}^\infty{\frac{1}{k!}} \leq \frac{1}{(n+1)!} ~~\Big[\leq \frac{1}{n!}\Big]$$  Is it possible to do this by induction? I don't really know how to approach this now.
115573
Let $n \in \mathbb{Z}^+$, prove the identity $$ \sum_{k=1}^{n-1} \binom {n} {k} \frac{kn^{n-k}}{k+1}=\frac{n(n^{n}-1)}{n+1}$$  First of all  $$ \sum_{k=1}^{n-1} \binom {n} {k} \frac{kn^{n-k}}{k+1}=n^{n}\Bigg(\sum_{k=1}^{n-1} \binom {n} {k} \frac{kn^{-k}}{k+1} \Bigg)$$ $$=n^n\Bigg(\sum_{k=1}^{n-1} \binom {n}{k} \bigg(1-\frac{1}{k+1}\bigg)\bigg(\frac{1}{n^k}\bigg)\Bigg)$$ $$=n^n \sum_{k=1}^{n-1} \binom {n} {k} \frac{1}{n^k}-n^n \sum_{k=1}^{n-1} \binom {n}{k}  \bigg( \frac{1}{k+1}\bigg)\bigg(\frac{1}{n^k} \bigg)$$  We have for the first sum $$(1+\frac{1}{x})^n = \sum_{k=0}^n \binom{n}{k}\frac{1}{x^k}.$$ For the second sum  $$(1+x)^n = \sum_{k=0}^n \binom{n}{k}x^k.$$  Integrating both sides from $0$ to $x$, we see that   $$\frac{(1+x)^{n+1}-1}{n+1} = \sum_{k=0}^n \binom{n}{k}\frac{x^{k+1}}{k+1}.$$  Putting $x=1$ yields $$\frac{2^{n+1}-1}{n+1}=\sum_{k=0}^n \binom{n}{k}\frac{1}{k+1}$$  Here where I have stopped. I could not get them similar for what I have. Would someone help me out !
67806
$$\frac{1}{n+1} + \frac{1}{n+2} + \cdots +\frac{1}{n+n} + \frac{1}{4n} &gt; \ln 2$$  $n$ is positive integer.  Thank you !
105627
I need to prove that $\left(1 + \frac 1 {n^2} \right)^n \to 1$.  I tried to use Bernoulli's inequality, but that is not very useful since in the original sequence there is a plus sign. I then tried to use the Sandwich Theorem by finding two sequences which would make bounds for the original one. The lower bound is obvious, the upper bound not so much. I tried using the sequence $\left( \frac 1 {n+1} \right) ^{\frac 1 n}$, but I could not show that this sequence is bigger than the original one for all $n$. Could anyone help me with this?
80322
Prove: $$ (a+b)^\frac{1}{n} \le a^\frac{1}{n} + b^\frac{1}{n}, \qquad \forall n \in \mathbb{N} $$ I have have tried using the triangle inequlity $ |a + b| \le |a| + |b| $,  without any success.
78732
Prove by mathematical induction that:  $$\forall n \in \mathbb{N} : \sum_{k=1}^{2n} \frac{(-1)^{k+1}}{k} = \sum_{k=n+1}^{2n} \frac{1}{k} $$  Step 1: Show that the statement is true for $n = 1$:  LHS = $$\frac{(-1)^{1+1}}{1} = 1$$  RHS = $$\frac{1}{1} = 1$$  Step 2: Show that "if true for n = p, then true for n = p + 1":  Starting with the LHS of equality for $n = p + 1$ and try to get to the RHS by using the equality for $n = p$. Simplifying:  $$\sum_{k=1}^{2(p+1)} \frac{(-1)^{k+1}}{k} = \sum_{k=1}^{2p+2} \frac{(-1)^{k+1}}{k}$$  Breaking out the first two term:  $$ \sum_{k=1}^{2p+2} \frac{(-1)^{k+1}}{k} = \frac{(-1)^{2p+3}}{2p+2} + \frac{(-1)^{2p+2}}{2p+1} + \sum_{k=1}^{2p} \frac{(-1)^{k+1}}{k} $$  The last term can now be exchanged for the RHS in the original equality:  $$\frac{(-1)^{2p+3}}{2p+2} + \frac{(-1)^{2p+2}}{2p+1} + \sum_{k=1}^{2p} \frac{(-1)^{k+1}}{k} = \frac{(-1)^{2p+3}}{2p+2} + \frac{(-1)^{2p+2}}{2p+1} + \sum_{k=p+1}^{2p} \frac{1}{k}$$  Since $2p+3$ is odd and $2p+2$ is even, we get:  $$\frac{(-1)}{2p+2} + \frac{1}{2p+1} + \sum_{k=p+1}^{2p} \frac{1}{k} = \frac{(-1)}{2p+2} + \sum_{k=p+1}^{2p+1} \frac{1}{k}$$  ...because we can easily absorb the second term in the third term sum. However, I am stuck here because the first term is negative. I probably made a trivial error somewhere, but I am unable to find it. Any suggestions?
50725
Prove that for $n\geq 2, \: \sum_{k = 1}^{n} \frac{1}{k^{2}} &lt; 2 - \frac{1}{n} $  I used induction and I compared the LHS and the RHS but I'm getting an incorrect inequality
4889
The question is to prove the inequality $$\frac{1}{n+1} \leq \ln \left(1+\frac 1n\right) \leq \frac 1n\\\forall n \geq 1, n\in \mathbb N$$  I tried using Taylor expansion but couldn't figure out anything. Any ideas? Thanks.