Using $$(a+b)^{n} =  \sum_{i=0}^n {n \choose k} a^{n-k} b^{K}$$  prove that   $$n^{1/n} - 1 \le \sqrt{\frac 2n}$$ for n= 2,3,4....   I know the first step is to set $$ n^{\frac 1n} = 1 + x $$ for some x>0 and then raise both sides to the n, but I'm lost after that.
123132
I am having difficulty proving the following inequality: $$ \sqrt[n]{n} &lt; 1 + \sqrt{\frac{2}{n}} \quad \text{for all positive integers}\,\,\, n. $$ I am trying to use mathematical induction but I am having trouble going from the left side to the right side (in the induction step) and vice versa.   I also tried to take the power of $n$ on both sides and somehow use the binomial expansion but I am having difficulty with that. Thank you!
58247
Let $n$ be a natural number and $a_1,a_2,\ldots,a_n$ are real numbers. Then prove that  $|\sum_{i=1}^n a_i|\le  \sqrt{n\cdot \sum_{i=1}^n a_i^2}$.   At first I tried to prove for $n=2$ i.e, $|a_1+a_2|\le \sqrt{2(a_1^2+a_2^2)}$, but unable to proceed. Give me some hint to proceed.
26124
Can someone look at my proof. I am supposed to prove by induction. The question is to prove the following:  $$\sum _{i=0}^{n}{i} =\frac { n\left( n+1 \right)  }{ 2 } .$$  If $n=1$  Then  $$1=\sum_{i=1}^1{i}=\frac{1(1+1)}{2}=1$$  Now assume $n=k$.  Thus  $$\sum_{i=1}^k{i}=\frac{k(k+1)}{2}=1$$  If now $n=k+1$  $$\sum_{i=1}^{k+1}{i}=\sum_{i=1}^k{i}+(k+1)= \frac{(k+1)(k+2)}{2}$$  Is there anything more that I have to do, or is this it?
37643
$$\sum_{k=0}^nq^k = \frac{1-q^{n+1}}{1-q}$$  I want to prove this by induction. Here's what I have.  $$\frac{1-q^{n+1}}{1-q} + q^{n+1} = \frac{1-q^{n+1}+q^{n+1}(1-q)}{1-q}$$  I wanted to factor a $q^{n+1}$ out of the second expression but that 1- is screwing it up...
88141
So I have to prove:      For each natural number greater than or equal to 3,   $$(1+\frac1n)^n&lt;n$$   My work: Basis step: $n=3$ $$\left(1+\frac13\right)^3&lt;3$$ $$\left(\frac43\right)^3&lt;3$$ $$\left(\frac{64}{27}\right)&lt;3$$ which is true.  Now the inductive step, assume $P(k)=\left(1+\frac1k\right)^k&lt;k$ to be true and prove $P(k+1)=\left(1+\frac1{k+1}\right)^{k+1}&lt;k+1$.   This is where I am stuck because usually you add or multiply by $k+1$ or some similar term.
12892
Proving that $$\sum_{n=0}^{\infty }\frac{1}{(2n)!!}=\sqrt{e}$$ Firstly, I tried to check the value with the exponential function at $x=.5$ but I found its terms not equal to the series terms.
60101
Let $n \in \mathbb N^+$. Show that if $x_1, x_2, ... , x_n$ are $n$ real numbers such that $-1 \le x_i \le 0$ for each $1 \le i \le n$, then $$(1 + x_1)(1 + x_2)...(1 + x_n) \ge 1 + x_1 + x_2 + ... + x_n$$  I am using a proof by induction, but I am unable to complete it. I would appreciate any tips on how I could finish the proof.   Incomplete Proof Let $P(n)$ be the proposition that $(1 + x_1)(1 + x_2)...(1 + x_n) \ge 1 + \sum_{i = 1} ^ n x_i$. When $n = 1$, LHS = RHS = $1 + x_i$. Clearly $1 + x_i \ge 1 + x_i$ and so $P(1)$ is true. Assume that $P(k)$ is true, i.e. $(1 + x_1)(1 + x_2)...(1 + x_k) \ge 1 + \sum_{i=1}^k x_i$. Case 1 Suppose $x_{k + 1}  = -1$. Then $1 + x_{k + 1} = 0$, hence LHS = $(1 + x_1)(1 + x_2)...(1 + x_k)(1 + x_{k+1}) = 0$. RHS = $1 + x_1 + x_2 + ... + x_k - 1 = x_1 + x_2 + ... + x_k$ which is always nonpositive since each $x_i$ is nonpositive. Hence RHS $\le 0$ and so LHS $\ge$ RHS. Case 2 Suppose $x_{k+1} = 0$. Then $1 + x_{k+1} = 1$, hence multiplying it to LHS does not change anything. Also, adding $x_{k+1}$ to RHS does not change anything. Therefore, LHS $\ge$ RHS still holds. Case 3 Suppose $-1 &lt; x_{k+1} &lt; 0$. Then $0 &lt; 1 + x_{k+1} &lt; 1$. Adding $x_{k+1}$ to RHS makes the RHS smaller since $x_{k+1}$ is negative. If LHS $= 0$, then multiplying $1 + x_{k+1}$ to LHS does not change anything, and so LHS $\ge$ RHS holds. If LHS > $0$, then multiplying $1 + x_{k+1}$ to LHS makes the LHS smaller since $0 &lt; 1 + x_{k+1} &lt; 1$. In this case LHS $\ge$ RHS does not necessarily hold? ...
123923
I'm trying to prove binomial theorem by induction, but I'm a little stuck. I would look at online resources as this problem has been done many times, but the version I am trying to prove the binomial theorem in a different form.  $$(1 + x)^n = \sum_{k = 0}^{n} \binom{n}{k} x^k$$  I'm mostly confused as to how I can make the left side be equivalent to a summation, any help is appreciated. Try to hint me along!
111590
I am trying to prove that $\binom{2n}{n} &lt; \frac{4^n}{\sqrt{2n}}$. I tried induction, but with no effect (all I can get to is $(2n+1)(2n+2) &lt; 4\sqrt{n(n+1)}$ which is false)
104983
$a _{1} = \frac{1}{2}, a _{2} = 1, a _{n}= \frac{1}{2} a_{n-1} + \sqrt{a _{n-2} }$  Show by the induction that it's $\le 4$ and also by the induction that $a_{n+1}-a_n \ge 0$  If it comes to first I don't know what to do and if it comes to second I tried many different ways but I couldn't prove it.
120175
I want to justificate following equation:  $$\sum_{k=0}^n \frac{(-1)^k}{k!(n-k)!}\frac{1}{2k+1} = \frac{2^n}{(2n+1)!!}$$  I calculated the both sides for $n$ from 1 to 10 and it was true. How the mathematical induction can be applied to this equation? Or is there other way to justificate it?
30701
$$\sum_{i=1}^n \frac1{n+i} = \sum_{i=1}^{2n}\frac{(-1)^{i+1}}i,\quad \text{ for }n\ge 1.$$  I really have no clue how on to solve this one. It's the last one on my assignment. Would really enjoy some help with this one.
26465
A teacher gave this as a homework question, and I have tried but haven't been able to arrive at a solution. $\sum_{k=0}^n {n+k \choose k} \frac{1}{2^{k}}= 2^{n}$ Could someone prove it, or at least point me in the right direction ?
73670
Knowing that $z + \frac{1}{z} = 3$ ($z \in \mathbb{R}$), prove that $z^n + \frac{1}{z^n} \in \mathbb{Q}$ ($n \in \mathbb{N}$ and $n \geq 1$).  I've tried to find a general formula for $z^n + \frac{1}{z^n}$. $$\left(z + \frac{1}{z}\right)^n = \binom{n}{0}z^n + \binom{n}{1}z^{n-2} + ... + \binom{n}{n-1}z^{2-n} + \binom{n}{n}z^{-n}$$ $$\left(z + \frac{1}{z}\right)^n = \binom{n}{0}(z^n + z^{-n}) + \binom{n}{1}(z^{n-2} + z^{2-n}) + ...$$  However, this formula seems to be wrong and I currently have no idea how to correct it and what to do next.  Thank you in advance for your help!
9005
I'm trying to solve  proposed by the department of mathematics of Oxford University. I don't know how to give a valid proof for the third one which says the following:     Prove that for $n \in \mathbb{N}^{\ge 1}$: $$ \sqrt{n} \le  \sum_{k=1}^n \frac{1}{\sqrt{k}} \le 2 \sqrt{n} - 1 $$   My first approach has been to convert the summation into a function in terms of $n$ so that I could find the maximum and minimum values in the interval. However, there doesn't seem to be an easy way to express the summation as a function. Thanks in advance!
332
I got quite some trouble trying to prove this.  $$\sum_{k=m}^{n}(-1)^k\binom{n}{k}\binom{k}{m}=0, n&gt;m\geq 0$$  I tried using $$\binom{n}{m}\binom{m}{k}=\binom{n}{k}\binom{n-k}{m-k}$$ and then changing the summation boundaries as $k=0, n=n-m$, but that took me nowhere... I would really appreciate help. Thank you.
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.