real analysis - To find the sum of the series $,1+ frac{1}{3cdot4}+frac{1}{5cdot4^2}+frac{1}{7cdot4^3}+ldots$

The answer given is $\log 3$.
Now looking at the series

\begin{align}
1+ \dfrac{1}{3\cdot4}+\dfrac{1}{5\cdot4^2}+\dfrac{1}{7\cdot4^3}+\ldots &=
\sum\limits_{i=0}^\infty \dfrac{1}{\left(2n-1\right)\cdot4^n}
\\
\log 3 &=\sum\limits_{i=1}^\infty \dfrac{\left(-1\right)^{n+1}\,2^n}{n}
\end{align}

How do I relate these two series?

Answer

Hint: a common series that is used for computing log of any real number is
$$
\log\left(\frac{1+x}{1-x}\right)=2\left(x+\frac{x^3}3+\frac{x^5}5+\frac{x^7}7+\dots\right)
$$
$u=\frac{1+x}{1-x}\iff x=\frac{u-1}{u+1}$

calculus - Evaluate the series $sum_{n=0}^{infty} frac{n^2-1}{n!}frac{x^n}{n-1}$

I would like to show that the following sum converges $\forall x \in \mathbb{R}$ as well as calculate the sum:

$\sum_{n=0}^{\infty} \frac{n^2-1}{n!}\frac{x^n}{n-1}$

First for the coefficient:

$\frac{n^2-1}{n!(n-1)}=\frac{n+1}{n!}$

Then, what I did was to try and formulate this series to a series which I know:

$\sum_{n=0}^{\infty} \frac{n^2-1}{n!}\frac{x^n}{n-1}=\sum_{n=0}^{\infty} \left( \frac{n+1}{n!} \right)x^n=\cdots = \frac{1}{x} \sum_{n=0}^{\infty}\frac{(n+1)^2 x^{n+1}}{(n+1)!}$

I have ended up with this formula, which reminds me somehow the expansion of the exponential $e^x$

$\sum_{n=0}^{\infty}\frac{x^n}{n!}=e^x$

but I cannot see how the term $(n+1)^2$ affects the result.

Thanks.

Answer

One should instead notice that

$$\frac{n+1}{n!}=\frac n{n!}+\frac1{n!}=\frac1{(n-1)!}+\frac1{n!}$$

And then we get the well-known series expansion for $e^x$.

elementary number theory - Prove that$gcd(a+b, a-b) = gcd(2a, a-b) = gcd(a+b, 2b) $

Question:

Prove that$\gcd(a+b, a-b) = \gcd(2a, a-b) = \gcd(a+b, 2b) $

My attempt:

First we prove $\gcd(a+b, a-b) = \gcd(2a, a-b)$.

Let $ d = \gcd(a+b, a-b)$ and $ \ e = \gcd(2a, a-b)$

$ d|a+b$ and $ \ d|a-b \implies \exists m,n \in \mathbb{Z}$ such that $ \ a+b = dm$ and $\ a-b = dn \implies a + a -dn = dm \implies 2a = dm + dn \implies 2a = d(m+n) \implies d|2a$

So, $ \ d|a-b$ and $ \ d |2a \implies d\le e$

$e|2a$ and $ \ e|a-b \implies \exists m,n \in \mathbb{Z}$ such that $ \ 2a = em$ and $\ a-b = en \implies 2a - (a-b) = a + b = em-en = e(m-n) \implies e|(a+b)$

So, $ \ e|(a-b)$ and $ \ e |(a+b) \implies e\le d$

Hence $ e =d$

Similarly, if I prove that $\gcd(a+b, a-b) = \gcd(a+b, 2b)$, will the proof be complete?

I am not quite sure if this is the correct way to prove the problem. My professor used a similar approach to prove that $ \ \gcd(a,b) = \gcd( a- kb, b)$.

Answer

Your approach is correct. However a few steps can be shortened.

Let $d=\gcd(a+b,a-b)$, then $d | a+b$ and $d | a-b$, thus $d$ divides all linear combinations of $a+b$ and $a-b$, in particular $d|(a+b)-(a-b)=2b$ and $d|(a+b)+(a-b)=2a$. Thus $d$ divides both $2a$ and $2b$.

Now you can take it from here.

abstract algebra - Polynomial with n real roots

Let $P(x) = x^n + a_{n-1}x^{n-1} + \cdots + a_{1}x + 1$ where $a_i$ are nonnegative and real. Assume $P$ has $n$ real roots.

Prove $P(2) \geq 3^n$.

I thought I had a good idea about rewriting $P$ as $(x-\alpha_1)\cdots(x-\alpha_n)$. The fact that the roots are real means that you can order them. Choose the largest root, $\alpha_k$, then $2-\alpha_k$ would have the smallest such value among the $\alpha_i$. And so we would have $P(2) \geq (2-\alpha_k)^n$.

But I haven't been able to think of a way to compare it with $3^n$.

Answer

The proof follows from the following Lemma:

Lemma If $0$$(2+a)(2+b) \geq 3(2+ab)$$

Proof:
$$(2+a)(2+b) \geq 3(2+ab) \Leftrightarrow \\
4+2a+2b+ab \geq 6+3ab \Leftrightarrow \\
0 \geq 2-2a-2b+2ab \Leftrightarrow \\
0 \geq 2(a-1)(b-1)
$$

QED Lemma

Now, lets solve the problem. Let $b_i=-\alpha_1$, and we can assume without loss of generality that $b_1 \leq b_2 \leq ... \leq b_n$.

As $b_1 \cdot .. \cdot b_n=1$ it follows that
$$b_n \geq 1 \\
b_1\cdot ... \cdot b_k \leq 1$$

Then, by repeadely applying the previous lemma, we get

$$(2+b_1)(2+b_2)(2+b_3)\cdot ... \cdot (2+b_n) \geq \\
3(2+b_1b_2)(2+b_3)\cdot ... \cdot (2+b_n) \geq \\
3^2(2+b_1b_2b_3)\cdot ... \cdot (2+b_n) \geq \\

....\\
3^{k-1}(2+b_1b_2b_3...b_k)(2+b_{k+1}\cdot ... \cdot (2+b_n) \geq \\
...\\
3^{n-1}(2+b_1b_2b_3...b_n)=3^n \\
$$

Much simpler second solution:

If $b_1,..,b_n$ are non-negative numbers then by the AM-GM inequality:

$$2+b_i = 1+1+b_i \geq 3 \sqrt[3]{b_i}$$

Therefore
$$(2+b_1)(2+b_2)(2+b_3)\cdot ... \cdot (2+b_n) \geq 3^n \sqrt[3]{b_1b_2...b_n}=3^n$$

Limit $lim_{ntoinfty} n^2left(sqrt{1+frac{1}{n}}+sqrt{1-frac{1}{n}}-2right)$

Greetings I am trying to solve $$\lim_{n\to\infty} n^2\left(\sqrt{1+\frac{1}{n}}+\sqrt{1-\frac{1}{n}}-2\right)$$ Using binomial series is pretty easy: $$\lim_{n\to\infty}n^2\left(1+\frac{1}{2n}-\frac{1}{8n^2}+\mathcal{O}\left(\frac{1}{n^3}\right)+1-\frac{1}{2n}-\frac{1}{8n^2}+\mathcal{O}\left(\frac{1}{n^3}\right)-2\right)=\lim_{n\to\infty}n^2\left(-\frac{1}{8n^2}+\mathcal{O}\left(\frac{1}{n^3}\right)-\frac{1}{8n^2}+\mathcal{O}\left(\frac{1}{n^3}\right)\right)=-\frac{1}{4}$$ The problem is that I need to solve this using only highschool tools, but I cant seem too take it down. My other try was to use L'Hospital rule but I feel like it just complicate things. Maybe there is even an elegant way, could you give me some help with this?

Answer

Hint: multiplying numerator and denominator by $\sqrt{1+1/n}+\sqrt{1-1/n}+2$ we get
$$2 n^2 \frac{\sqrt{1-1/n^2}-1}{\sqrt{1+1/n}+\sqrt{1-1/n}+2}$$ and then do the same with $$\sqrt{1-1/n^2}+1$$
you will get

$$\frac{n^2(2(\sqrt{1-1/n^2}-1))(\sqrt{1-1/n^2}+1)}{(\sqrt{1+1/n}+\sqrt{1-1/n}+2)(\sqrt{1-1/n^2}+1)}$$

elementary number theory - What is the largest power of 2 that divides $200!/100!$.

What is the largest power of 2 that divides $200!/100!$.

No use of calculator is allowed.
I had proceeded in a brute force method which i know regret..
I would like to know your methods.

Answer

Find highest power of $2$ in $200!$ and $100!$, using Legendre's formula

In $200!$, highest power of $2$

$$=\lfloor 200/2 \rfloor +\lfloor 200/4 \rfloor +\lfloor 200/8 \rfloor +\lfloor 200/16 \rfloor +\lfloor 200/32 \rfloor +\lfloor 200/64 \rfloor +\lfloor 200/128 \rfloor $$

$$=100+50+25+12+6+3+1=197$$

In $100!$, highest power of $2$

$$=\lfloor 100/2 \rfloor +\lfloor 100/4 \rfloor +\lfloor 100/8 \rfloor +\lfloor 100/16 \rfloor +\lfloor 100/32 \rfloor +\lfloor 100/64 \rfloor$$

$$= 50 + 25+12+6+3+1 =97$$

Now, just subtract the two, and we get $100$ as the answer.

What 's the short proof that for square matrices $AB = I$ implies $BA = I$?

I'm trying to remember the one line proof that for square matrices $AB = I$ implies $BA = I$.
I think it uses only elementary matrix properties and nothing else. Does anyone know the proof?

I remember it beginning with $BAB = B$ and the result following almost immediately.