Let we have the following sequences $x_n=1/1^2 + 1/2^2 + ....1/n^2$ Prove that for any $n$ $x_n <= 2-1/n $ how can I calculate its limit ?
$y_n= \frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+..........+\frac{1}{\ \sqrt{n^2+n}}$ find the limit of the sequence $y_n$
Let $z_0=3$ and $ z_{n+1}=\frac{2}{z_n +1} $ and $t_n=\frac{z_n-1}{z_n+2}$ prove that $t_n$ is a geometric sequence find its general form
Please help me
For the first inequality we can use that $$\frac{1}{i^2}\le \frac{1}{i(i-1)}$$ for $$i>1$$
For the second limit, one can see that
$$\int_1^{n+1}\frac1{\sqrt{n^2+x}}\ dx\le\sum_{k=1}^n\frac1{\sqrt{n^2+k}}\le\frac1{\sqrt{n^2+1}}+\int_1^n\frac1{\sqrt{n^2+x}}\ dx$$
Evaluating these, we have
$$2(\sqrt{n^2+n+1}-\sqrt{n^2+1})\le\sum_{k=1}^n\frac1{\sqrt{n^2+k}}\le\frac1{\sqrt{n^2+1}}+2(\sqrt{n^2+n}-\sqrt{n^2+1})$$
As $n\to\infty$, we have, by squeeze theorem with a touch of binomial expanding,
$$y=\lim_{n\to\infty}\sum_{k=1}^n\frac1{\sqrt{n^2+k}}=1$$