It basically says it all in the title. I tried solving the inequality using the bernoulli inequality somehow $$\dfrac{\displaystyle\sum_{k=0}^n\frac{1}{k!}}{(1+\frac{1}{n})^n}\geq 1,$$ but the factorial gets me everytime.  Also I've tried induction without success:  $\displaystyle\sum_{k=0}^{n+1}\frac{1}{k!}\stackrel{IV}\geq \left(1+\frac{1}{n}\right)^n+\frac{1}{(n+1)!}$
122525
Prove that $$\left(1+\frac{1}{n}\right)^n&lt;\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+...+\frac{1}{n!}$$ for $n&gt;1 , n \in  \mathbb{N}$.
73413
I want to prove the following equality:  $$\displaystyle\sum_{k=1}^{n}k{n \choose k}2^{n-k}=n3^{n-1}$$  So I had an idea to use $((1+x)^n)'=n(1+x)^{n-1}$  So I could just use the binomial theorem and let $$x=2 \implies (n3^{n-1})$$ and then modify the sum into that one on the left side.  So I need to prove that : $$\displaystyle\sum_{k=1}^{n}k{n \choose k}2^{n-k}=n(1+x)^{n-1}$$ if$$x=2$$ Any help would be appreciated.
54720
Mathematica tells me that  $$\sum _{k=0}^n { n \choose k} \frac{(-1)^k}{2k+1} = \frac{(2n)!!}{(2n+1)!!}.$$  Although I have not been able to come up with a proof.  Proofs, hints, or references are all welcome.
16144
I am trying to prove the summation formula using induction:  $$\sum_{k=1}^n \frac{1}{k(k+1)} = 1-\frac{1}{n+1}$$  So far I have...  Base case:   Let n=1 and test   $\frac{1}{k(k+1)} = 1-\frac{1}{n+1}$  $\frac{1}{1(1+1)} = 1-\frac{1}{1+1}$    $\frac{1}{2} = \frac{1}{2}$   True for n=1   Induction Hypothesis:   Assume the statement is true for the n-th case   $\sum_{k=1}^n  \frac{1}{k(k+1)} = 1-\frac{1}{n+1}$  Inductive Step:   Prove, using the Inductive Hypothesis as a premise, that   $$\sum_{k=1}^{n+1}  \frac{1}{k(k+1)} = \sum_{k=1}^{n}  \frac{1}{k(k+1)} + \frac{1}{(n+1)(n+2)} = 1-\frac1{n+1} + \frac{1}{(n+1)(n+2)} = \frac{(n+1)(n+2)}{(n+1)(n+2)}+\frac{-2-n}{(n+1)(n+2)}+\frac{1}{(n+1)(n+2)} = \frac{(n+1)(n+2)-2-n+1)}{(n+1)(n+2)} = \frac{(n+1)(n+2)-n-1}{(n+1)(n+2)} = \frac{n^2+2n+1}{(n+1)(n+2)} = \frac{(n+1)(n+1)}{(n+1)(n+2)} = \frac{n+1}{n+2}$$  To prove $$ 1-\frac{1}{n+2} = \frac{n+1}{n+2} $$ Multiply both sides by $n+2$ to get an equivalent expression. $$ (1-\frac{1}{n+2}) * (n+2) = (\frac{n+1}{n+2}) * (n+2)  $$ $$ n+1=n+2−1 $$  Does this all make sense? How can this be improved upon?
10799
The simple inequality that $\left(\frac{n}{k}\right)^k \leq \binom{n}{k}$ has a number of different proofs. But is there a particularly intuitive, short and elegant proof that uses the natural interpretation of binomial coefficients, for example. I would ideally like a proof which is also accessible to students with very limited prerequisite knowledge.  Here is the best proof that I have seen which is less intuitive than I was hoping for.  First we first prove that $$\frac{n-i}{k-i} \geq \frac{n}{k}$$  for $i&lt;k\leq n$.  This follows from  $$0\leq (n-k)i = k(n-i) - n(k-i) =  k(k-i)\left(\frac{n-i}{k-i}-\frac{n}{k}\right),$$ and $k(k-i)&gt; 0$, so $(n-i)/(k-i) \geq n/k.$  Now we multiply the over $i\in\{1,\ldots,k-1\}$ to obtain  $$\frac{n(n-1)\cdots(n-k+1)}{k(k-1)\cdots 1} \geq \frac{n^k}{k^k},$$ or equivalently $\binom{n}{k}\geq (n/k)^k$.
119945
I'm about to prove that for any $a,b&gt;0$ and $n\in\mathbb{N},$ the inequality: $\frac{a^n+b^n}{2}\geq\left(\frac{a+b}{2}\right)^n$ holds.   By induction I get: $$\left(\frac{a+b}{2}\right)\cdot\frac{a^n+b^n}{2}$$ $$=\frac{1}{2}\cdot\frac{a^{n+1}+b^{n+1}+ab^n+ba^n}{2}$$ $$=\frac{1}{2}\left(\frac{a^{n+1}+b^{n+1}}{2}+\frac{ab^n+ba^n}{2}\right).$$  Now I have to prove that $$\frac{1}{2}\left(\frac{a^{n+1}+b^{n+1}}{2}+\frac{ab^n+ba^n}{2}\right)$$ $$\leq\frac{a^{n+1}+b^{n+1}}{2}$$ $$\Leftrightarrow a^{n+1}+b^{n+1}\geq ab^n +ba^n.$$ But I don't know how to prove that one, is it possible to be proved by induction ? Thank you.
62299
I have to show $n! \leq \left( \frac{n+1}{2} \right)^n$ via induction.  This is where I am stuck:  $$\left( \frac{n+2}{2} \right)^{n+1}  \geq \dots \geq =2  \left( \frac{n+1}{2} \right)^{n+1} = \left( \frac{n+1}{2} \right)^n(n+1) \geq n!(n+1) = (n+1)! $$  I approached this from both sides and this is the closest I can get. I realize that $n+2$ on the left has to be bigger than $n+1$ on the right, but I do not know who to show that it overpowers the factor two I have from the right.  What could I do to fill the dots? Currently, I just have it without the dots, but I would be happier if I could back it up.
41292
I would appreciate some help proving the inequality $$\sum_{k=n+1}^{\infty}\frac{1}{k^{2}\log k}\leq\frac{1}{n\log n}.$$ Thanks in advance!
76504
I'm struggling with proving the following inequality: $$\sum_{i=0}^{t-2}{\frac{1}{t+3i}} \leq \frac{1}{2}$$  for all $t \geq 2$. I think it is monotonic non-increasing in $t$, which would suffice.  Thanks in advance!
85800
I'm having trouble proving this by induction. We need to show that $P(k+1)$ is true: $$\sum_{i=2}^{k+1} \frac{1}{i^2}\leq 1.$$  Don't know where to go from here. Any help?
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.
41068
Problem is to find sum  $$\frac{1}{3!}+\frac{4}{4!}+\frac{9}{5!}\cdots$$  What I knew doesn't apply on this problem  Some series are telescoping, some types are solvable using binomial , both look useless here Binomial gives  $$n (n-1) (n-2)x^3=1$$  $$n (n-1)( n-2)(n-3)x^4=4$$  What approach to use here?
365
How could we prove that this inequality holds      $$ \left(1+\frac{1}{n+1}\right)^{n+1} \gt \left(1+\frac{1}{n} \right)^{n} $$   where $n \in \mathbb{N}$, I think we could use the AM-GM inequality for this but not getting how?
88351
I would like to prove the following inequality, for $n=0,1,2,...$,   $$ \binom{2n}{n}\leq 4^n.$$ I already proved it by induction, and I'm looking for another proof.
100263
I'm trying to show that      $$\sum_{k=0}^n (-1)^k\frac{{ {n}\choose{k}}}{{ {x+k}\choose{k}}} = \frac{x}{x+n}$$   I've tried expanding this using definition of binomial coefficient, then binomial expansion, now I'm using induction to simplify $\sum_{k=0}^m (-1)^k\frac{{ {n}\choose{k}}}{{ {x+k}\choose{k}}}$, but it looks very inelegant, is there a better way?
41323
Prove this inequality, if $x_1^2+x_2^2+x_3^2=1$: $$ \sum_{i=1}^{3}\frac{x_i}{1+x_i^2} \le \frac{3\sqrt{3}}{4} $$  So far I got to $x_1^4+x_2^4+x_3^4\ge\frac{1}3$ by using QM-AM for $(2x_1^2+x_2^2, 2x_2^2+x_3^2, 2x_3^2+x_1^2)$, but to be honest I'm not sure if that's helpful at all.  (You might want to "rename" those to $x,y,z$ to make writing easier): $$ \frac{x}{1+x^2}+\frac{y}{1+y^2}+\frac{z}{1+z^2} \le \frac{3\sqrt3}4$$
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.