convergence of $\sum_{n=1}^{\infty} \frac{1}{(n+1)^2 -1}$

Question

How can i check convergence of $$\sum_{n=1}^{\infty} \frac{1}{(n+1)^2 -1}$$ ?

How don't sure how to check it. I tried some of the tests.

Answer 1

EG you can use comparison test with $\sum \frac{1}{n^2}$

Answer 2

$a_n:= \dfrac{1}{(n+1)^2 -1}=$

$\dfrac{1}{(n+2)n} \lt \dfrac{1}{n^2} :=b_n$

$\sum_{k=1}^{\infty} b_n$ converges.

You may want to look at:

Does sequence ${ s }_{ n }=\sum _{ k=0 }^{ n }{ \frac { 1 }{ { 1+k }^{ 2 } } } $ converge

Answer 3

You could use the comparison test with $\sum \frac1{n^2}$, or, for a more ‘elementary’ method, observe that,

\begin{align} \sum_{n=1}^{K} \frac2{(n+1)^2 - 1} =\,& \sum_{n=1}^{K} \frac2{n(n+2)}\\ =& \,\sum_{n=1}^{K} \left( \frac1{n} - \frac1{n+2} \right)\\[9pt] =& \,\left(\frac11 - \frac13\right) + \left(\frac12 - \frac14\right)\\ +&\,\left(\frac13 - \frac15\right) + \left(\frac14 - \frac16\right)\\[1pt] +& \quad \quad \, ...\\[1pt] +&\,\left(\frac1{K-1} -\frac1{K+1}\right) + \left(\frac1{K} - \frac1{K+2}\right)\\[9pt] =\,& \frac32 - \frac{2K + 3}{(K+1)(K+2)} \quad \text{(by telescoping)}\\ \rightarrow & \, \frac32 \quad \text{as K $\rightarrow \infty$} \end{align}

Another way to show that the sum converges is by the integral test (as usual with these types of questions, there are many ways to prove convergence)...