Linked Questions

18
votes
3answers
786 views

Find $\lim\limits_{n \to \infty}\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\cdots\left(1-\frac{1}{n^2}\right)$ [duplicate]

Find the limit $$\lim_{n \to \infty}\left[\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\cdots\left(1-\frac{1}{n^2}\right)\right]$$ I take log and get $$\lim_{n \to \infty}\sum_{k=2}^{n} \...
10
votes
4answers
488 views

Is $\prod \limits_{n=2}^{\infty}(1-\frac{1}{n^2})=1$ [duplicate]

Question is to check if $\prod \limits_{n=2}^{\infty}(1-\frac{1}{n^2})=1$ we have $\prod \limits_{n=2}^{\infty}(1-\frac{1}{n^2})=\prod \limits_{n=2}^{\infty}(\frac{n^2-1}{n^2})=\prod \limits_{n=2}^{\...
1
vote
2answers
226 views

Limit $\lim_{n\to+\infty} (1-\frac1{2^2})(1-\frac1{3^2})\cdot \cdots \cdot(1-\frac{1}{n^2})$ and series $\sum_{n=2}^{\infty} \ln(1-\frac1{n^2})$ [duplicate]

Possible Duplicate: Finding Value of the Infinite Product $\prod \Bigl(1-\frac{1}{n^{2}}\Bigr)$ Compute: \begin{align*} \lim_{n\to+\infty} (1-\frac{1}{2^2})(1-\frac{1}{3^2})(1-\frac{1}{4^2})\...
3
votes
3answers
133 views

Limit Involving a Product of a Sequence: $\lim _{x\to\infty }\left(1-\frac1{2^2}\right)\left(1-\frac1{3^2}\right)\dots\left(1-\frac1{x^2}\right)$ [duplicate]

I am having trouble figuring out how to solve this limit. $$\lim _{x\to \infty }\left(\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)...\left(1-\frac{1}{x^2}\right)\right)$$ I understand ...
1
vote
1answer
127 views

Calculate $\prod_{n=2}^\infty\bigg(1-\frac1{n^2}\bigg)$ [duplicate]

How do you compute $\prod_{n=2}^\infty\bigg(1-\frac1{n^2}\bigg)$ I have no clue on how to start. Any help would be appreciated.
14
votes
6answers
5k views

The sum of series with natural logarithm: $\sum_{n=1}^\infty \ln\left(\frac{n(n+2)}{(n+1)^2}\right)$ [duplicate]

Calculate the sum of series: $$\sum_{n=1}^\infty \ln\left(\frac{n(n+2)}{(n+1)^2}\right)$$ I tried to spread this logarithm, but I'm not seeing any method for this exercise.
5
votes
3answers
628 views

how to evaluate the product $\prod _{n=2}^\infty (1+ \frac{1}{n^2}+\frac{1}{n^4}+\frac{1}{n^6}+\cdots )$? [closed]

Evaluating the infinite product of $\prod _{n=2}^\infty (1+ \frac{1}{n^2}+\frac{1}{n^4}+\frac{1}{n^6}+\cdots )$. Please Help.
11
votes
2answers
3k views

How was Euler able to create an infinite product for sinc by using its roots?

In the Wikipedia page for the Basel problem, it says that Euler, in his proof, found that $$\begin{align*} \frac{\sin(x)}{x} &= \left(1 - \frac{x}{\pi}\right)\left(1 + \frac{x}{\pi}\right)\left(1 ...
8
votes
1answer
1k views

Trying to get the infinite product for $\sin x$

I start with the fact that the zeros of $\sin x$ are $ n\pi$, $n\in\mathbb{Z}$. Therefore, it should be possible to express it as an infinite product: $$\sin x = x (x-\pi)(x+\pi)(x-2\pi)(x+2\pi)\...
6
votes
1answer
1k views

How to evaluate this infinite product: $\prod\limits_{n=1}^{\infty }{\left( 1-\frac{1}{{{2}^{n}}} \right)}$ [duplicate]

How to evaluate this one $$\prod\limits_{n=1}^{\infty }{\left( 1-\frac{1}{{{2}^{n}}} \right)}$$
13
votes
0answers
622 views

On the weak and strong convergence of an iterative sequence

I have some difficulties in the following problem. I would like to thank for all kind help and construction. Let $H$ be an infinite dimensional real Hilbert space and $F: H\rightarrow H$ be a ...
1
vote
3answers
82 views

Find the value of the Product $\prod_{k=1}^{\infty} \frac{2k(2k+2)}{(2k+1)^2}$

Find the value of the Product $$P=\prod_{k=1}^{\infty} \frac{2k(2k+2)}{(2k+1)^2}$$ we have $$P=\prod_{k=1}^{\infty}\left(1-\frac{1}{(2k+1)^2}\right)$$ Taking $ln$ on both sides we get $$\ln(P)=\...
3
votes
1answer
452 views

Convergence of the infinite product $\prod_{n=1}^{\infty}\frac{1}{n^{2}+1}$

Does this product converge? $$\prod_{n=1}^{\infty}\frac{1}{n^{2}+1}$$ any hint?
6
votes
1answer
207 views

Limit of $\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^3}\right)\dots \left(1-\frac{1}{n^n}\right)$ as $n\to \infty$

So I'm trying to solve the following limit: $$\lim_{n \to \infty}\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^3}\right)\dots \left(1-\frac{1}{n^n}\right)$$ Now, I tried getting the squeeze ...
3
votes
3answers
90 views

how to prove that this sequence converges to $0$?

$0 \le x_0 \le \frac{1}{2}$ , and $x_{n+1}=x_n-\dfrac{4x_n^3}{n+1}$ When I take $x_0=\sqrt{\frac{1}{12}}$, it converges very very slow. I can see it is monotonic decreasing but don't know how to ...

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