# Thread: discuss the convergence of the series

1. ## discuss the convergence of the series

$[(1+\displaystyle \frac{1}{2})^2- (1+\displaystyle \frac{1}{3})^2]+[(1+\displaystyle \frac{1}{4})^2-(1+\displaystyle \frac{1}{5})^2]+...+[(1+\displaystyle \frac{1}{2n})^2 - (1+\displaystyle \frac{1}{2n+1})^2]+...$

What is the effect of removing the square brackets?

does removing the square brackets mean remove ()^2?

2. Originally Posted by wopashui
$[(1+\displaystyle \frac{1}{2})^2- (1+\displaystyle \frac{1}{3})^2]+[(1+\displaystyle \frac{1}{4})^2-(1+\displaystyle \frac{1}{5})^2]+...+[(1+\displaystyle \frac{1}{2n})^2 - (1+\displaystyle \frac{1}{2n+1})^2]+...$

What is the effect of removing the square brackets?

does removing the square brackets mean remove ()^2?
Removing the square brackets means that the general terms $(-1)^{n} (1+\frac{1}{n})^{2}$ doesn't tend to 0 if n tends to infinity...

Kind regards

$\chi$ $\sigma$

3. Originally Posted by chisigma
Removing the square brackets means that the general terms $(-1)^{n} (1+\frac{1}{n})^{2}$ doesn't tend to 0 if n tends to infinity...

Kind regards

$\chi$ $\sigma$

so does the series converge to 0 at the beginning? then diverge after removing the bracket?

4. Originally Posted by wopashui
so does the series converge to 0 at the beginning? then diverge after removing the bracket?
Do you agree that $\displaystyle\sum\limits_{k = 2}^\infty {( - 1)^k \left( {1 + \frac{1}{k}} \right)^2 }$ is the same as you posted?

If not, why not?

If yes, for any series $\sum\limits_{n = k}^\infty {a_n }$ if $(a_n)\not\to 0$ then the series diverges.
This is known as the first divergent test.