Knowing that $a_n \rightarrow a$ and $b_n=\frac{a_1+a_2+...+a_n}{n}$. How can I show that $b_n \rightarrow a$?
4438
Prove that if $\lim_{n \to \infty}z_{n}=A$ then:  $$\lim_{n \to \infty}\frac{z_{1}+z_{2}+\cdots + z_{n}}{n}=A$$ I was thinking spliting it in: $$(z_{1}+z_{2}+\cdots+z_{N-1})+(z_{N}+z_{N+1}+\cdots+z_{n})$$ where $N$ is value of $n$ for which $|A-z_{n}|&lt;\epsilon$ then taking the limit of this sum devided by $n$ , and noting that the second sum is as close as you wish to $nA$ while the first is as close as you wish to $0$. Not sure if this helps....
93582
Let $a,b,$ and $c$ be positive numbers with $a+b+c = 1$. Prove that $$\left (\dfrac{1}{a}+1 \right)\left (\dfrac{1}{b}+1 \right)\left (\dfrac{1}{c}+1 \right) \geq 64.$$   Attempt  Expanding the LHS we obtain $\left (\dfrac{1+a}{a} \right)\left (\dfrac{1+b}{b} \right)\left (\dfrac{1+c}{c} \right)$. We are given that $a+b+c = 1$, so substituting that in we get $\left (\dfrac{b+c+2a}{a} \right)\left (\dfrac{a+c+2b}{b} \right)\left (\dfrac{a+b+2c}{c} \right)$. Then do I say $\left (\dfrac{b+c+2a}{a} \right)\left (\dfrac{a+c+2b}{b} \right)\left (\dfrac{a+b+2c}{c} \right) \geq 64$ and see if I can get a true statement from this?
94783
How can I prove that $\sum^n_{i=1} \frac{1}{i(i+1)} = \frac{n}{n+1}$? I noticed that in the sum, the denominator has terms that cancel out, but I'm not sure how to take advantage of that.
63022
Given that $a \equiv b \pmod n$ and $c\equiv d \pmod n$, I need to prove that $ac \equiv bd \pmod n$  So far, I've only managed to deduce that $a+b \equiv c+d \pmod n$. I don't know if this is usable, but it's there, at least.  Any hint would be greatly appreciated!
97685
Consider the following sequence:     $a_{n+2} = \frac{a_{n+1}+a_{n}}{2}$, $a_{1}$ and $a_{2}$ are given.   Write $a_{n}$ as a function of $a_{1}$ and $a_{2}$ and show that its limit is $\frac{a1 + 2a_{2}}{3}$  I think I am loosing myself on algebra here. I can't even do the first part. Any help is welcome, thanks a lot!
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.
105434
Good morning, i have a big problem with this proof. I haven't idea about this proof.   Problem:   Suppose $\lim_{n\rightarrow\infty}a_n =L  $ then $\lim_{n\rightarrow\infty}\ \frac{a_1+a_2+\cdots+a_n}n=L  $  I've tried this:  $| a_n-L|&lt;\varepsilon\Rightarrow L-\varepsilon&lt;a_n&lt;L+\varepsilon$ Please help!!
94173
It's easy to check that for any natural $n$ $$\frac{n+1}{n}=\cfrac{1}{2-\cfrac{n+2}{n+1}}.$$  Now,   $$1=\frac{1}{2-1}=\frac{1}{2-\cfrac{1}{2-1}}=\frac{1}{2-\cfrac{1}{2-\cfrac{1}{2-1}}}=\cfrac{1}{2-\cfrac{1}{2-\frac{1}{2-\frac{1}{2-1}}}}=\ldots =\cfrac{1}{2-\cfrac{1}{2-\frac{1}{2-\frac{1}{2-\frac{1}{2-\dots}}}}},$$  $$2=\cfrac{1}{2-\cfrac{3}{2}}=\cfrac{1}{2-\cfrac{1}{2-\cfrac{4}{3}}}=\cfrac{1}{2-\cfrac{1}{2-\frac{1}{2-\frac{5}{4}}}}=\cfrac{1}{2-\cfrac{1}{2-\cfrac{1}{2-\frac{1}{2-\frac{6}{5}}}}}=\ldots =\cfrac{1}{2-\frac{1}{2-\frac{1}{2-\frac{1}{2-\frac{1}{2-\ldots}}}}}.$$  Since the right hand sides are the same, hence $1=2$.
118369
$$\sum_{i=0}^{n-1} ar^i = \frac{a(r^n-1)}{r-1}$$ for  $r \in \mathbb R, r  \ne 1, n \ge 1 $  i. Show true for $n=1$ $$\frac{a(r^1-1)}{r-1} = a$$  ii. I am stuck on how to do the second part of the proof, and the logic behind it.
65053
let sequence $\{a_{n}\}$ such $$a_{1}=1,a_{n+1}=1+\dfrac{n}{a_{n}}$$  show that: $$a_{n}=\sqrt{n}+\dfrac{1}{2}-\dfrac{1}{8\sqrt{n}}+o\left(\dfrac{1}{\sqrt{n}}\right)?$$
3540
I don't know what to do with this. Nothing works. I hope somebody can help me to find a decision $$n!\leq {\left(\frac{n+1}{2}\right)}^{n}$$
87175
So I start with:  $$a_n = \left(\frac23\right)^n + n$$ I know $a_1=\frac53, a_2=\frac{22}9, a_3=\frac{340}{81}, a_4=\frac{1247}{243} $ Then I do: $$a_{n-1} = \left(\frac23\right)^{n-1} + (n-1)$$    Do I now do $a_n - a_{n-1}$ or $a_n\over a_{n-1}$ ?
60943
If $a,b,c &gt; 0$ are such that $abc=1$, then   $$ \frac{a}{b} + \frac{b}{c} + \frac{c}{a} \geq a + b+ c. $$   I would be pleased if you give me a hint. Thanks in advance.
69440
Show by induction that for all natural numbers n>3  $(1+\frac{1}{n})^n&lt;n$  Let $(1+\frac{1}{n})^n&lt;n$ be true !   We show that $(1+\frac{1}{n+1})^{n+1})&lt;n+1$  $(1+\frac{1}{n})^n(1+\frac{1}{n})&lt;n(1+\frac{1}{n})$   $(1+\frac{1}{n})^{n+1}&lt;n+1$  So now we just need to show that   $(1+\frac{1}{n+1})^{n+1}&lt;(1+\frac{1}{n})^{n+1}$  So here , If I show that :   $(1+\frac{1}{n+1})&lt;(1+\frac{1}{n})$   Does that mean im done with exercise ?
72090
Show that $z\bar{a}+\bar{z}a \leq 2|a||z|$  I was able to check this using some examples which is not ideal for a mathematician way of proving a problem/case/theorem. I couldn't generalize it (prove it). I will be glad if I can be given the hint/proof for this problem. Thank you.
116766
How would we go about proving that $$\frac{1}{1\cdot 2} + \frac{1}{2\cdot 3} + \frac{1}{3 \cdot 4} +\ldots +\frac{1}{n(n+1)} = \frac{n}{n+1}$$
8061
Show that $\frac{1}{(n+1)!}(1+\frac{1}{(n+1)}+\frac{1}{(n+1)^2}+\frac{1}{(n+1)^3}+\cdots)=\frac{1}{n!n}$, (here $n$ is a natural number) Maybe easy, but I cannot see it. Thanks in advance! Alexander