# All Questions

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### How can I evaluate $\sum_{n=0}^\infty (n+1)x^n$

How can I evaluate $$\sum_{n=1}^\infty \frac{2n}{3^{n+1}}$$ I know the answer thanks to Wolfram Alpha, but I'm more concerned with how I can derive that answer. It cites tests to prove that it is ...
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### Why does $1+2+3+\dots = -\frac{1}{12}$?

$\displaystyle\sum_{n=1}^\infty \frac{1}{n^s}$ only converges to $\zeta(s)$ if $\text{Re}(s) > 1$. Why should analytically continuing to $\zeta(-1)$ give the right answer?
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### Is $.999999999… = 1$?

I'm told by smart people that $0.999999999\ldots = 1$ and I believe them, but is there a proof that explains why this is?
9k views

### How to prove that $\lim\limits_{x\to0}\frac{\sin x}x=1$?

How can one prove the statement $$\lim\limits_{x\to 0}\frac{\sin x}x=1$$ without using the Taylor series of $\sin$, $\cos$ and $\tan$? Best would be a geometrical solution. This is homework. In my ...
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### How to use the Extended Euclidean Algorithm manually?

I've only found a recursive algorithm of the extended Euclidean algorithm. I'd like to know how to use it by hand. Any idea?
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### Is $\frac{\textrm{d}y}{\textrm{d}x}$ not a ratio?

In the book Thomas's Calculus (11th edition) it is mentioned (Section 3.8 pg 225) that the derivative $\frac{\textrm{d}y}{\textrm{d}x}$ is not a ratio. Couldn't it be interpreted as a ratio, because ...
24k views

### Different methods to compute $\sum\limits_{n=1}^\infty \frac{1}{n^2}$

As I have heard people did not trust Euler when he first discovered the formula $$\zeta(2)=\sum_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{6}.$$ However, Euler was Euler and he gave other proofs. I ...
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### Prove that $\gcd(a^n - 1, a^m - 1) = a^{\gcd(n, m)} - 1$

For all $a, m, n \in \mathbb{Z}^+$, $$\gcd(a^n - 1, a^m - 1) = a^{\gcd(n, m)} - 1$$
9k views

### Value of $\sum\limits_n x^n$

Why is $\displaystyle \sum\limits_{n=0}^{\infty} 0.7^n$ equal $1/(1-0.7) = 10/3$ ? Can we generalize the above to $\displaystyle \sum_{n=0}^{\infty} x^n = \frac{1}{1-x}$ ? Are there some ...
9k views

### Zero to the zero power - Is $0^0=1$?

Could someone provide me with good explanation of why $0^0 = 1$? My train of thought: $x > 0$ $0^x = 0^{x-0} = 0^x/0^0$, so $0^0 = 0^x/0^x = ?$ Possible answers: $0^0 * 0^x = 1 * 0^x$, so ...
24k views

### Is value of $\pi = 4$?

What is wrong with this? SOURCE
8k views

### How can you prove that a function has no closed form integral?

I've come across statements in the past along the lines of "function $f(x)$ has no closed form integral", which I assume means that there is no combination of the operations: addition/subtraction ...
9k views

### $\sqrt a$ is either an integer or an irrational number.

I got this interesting question in my mind: How do we prove that if $a \in \mathbb N$, then $\sqrt a$ is an integer or an irrational number? Can we extend this result? That is, can it be shown ...
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### Examples of bijective map from $\mathbb{R}^3\rightarrow \mathbb{R}$

Could any one give an example of a bijective map from $\mathbb{R}^3\rightarrow \mathbb{R}$? Thank you.
15k views

### Proof for formula for sum of sequence $1+2+3+\ldots+n$?

Apparently $1+2+3+4+\ldots+n = \dfrac{n\times(n+1)}2$. How? What's the proof? Or maybe it is self apparent just looking at the above? Does this problem have a name and maybe a presence on the net? ...
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### $\sqrt{c+\sqrt{c+\sqrt{c+\cdots}}}$, or the limit of the sequence $x_{n+1} = \sqrt{c+x_n}$

(Fitzpatrick Advanced Calculus 2e, Sec. 2.4 #12) For $c \gt 0$, consider the quadratic equation $x^2 - x - c = 0, x > 0$. Define the sequence $\{x_n\}$ recursively by fixing $|x_1| \lt c$ and ...
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### If $a \mid m$ and $(a + 1) \mid m$, prove $a(a + 1) | m$.

Can anyone help me out here? Can't seem to find the right rules of divisibility to show this: If $a \mid m$ and $(a + 1) \mid m$, then $a(a + 1) \mid m$.
### Proof that $\sum\limits_{k=1}^nk^2 = \frac{n(n+1)(2n+1)}{6}$?
I am just starting into calculus and I have a question about the following statement I encountered while learning about definite integrals: $$\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$$ I really ...