What is the relation between sum? Is something interesting? Is it possible taking square or solving set of equation calculate "non squared" sum? Thank you for your explanation.
$$\sum\limits_{n=2}^{\infty}(\frac{1}{n}-\frac{1}{n-1})=-1$$
and
$$\sum\limits_{n=2}^{\infty}(\frac{1}{n^2}-\frac{1}{(n-1)^2})=\frac{\pi^2}{6}-1$$
On
In fact we have $$\sum_{n=1}^{\infty}a_{n+1}-a_n=(a_2-a_1)+(a_3-a_2)+(a_4-a_3)+\cdots=a_{\infty}-a_1$$ These interesting sequences are referred to as Telescoping series. Also the second series has been calculated wrong.
$$\sum\limits_{n=2}^{\infty}(\frac{1}{n^2}-\frac{1}{(n-1)^2})=-1$$
I do not understand much what you mean but maybe these points help you in something
1)The last result is wrong, first think that you can separate this sum into two parts $$\sum\limits_{n=2}^{\infty}(\frac{1}{n^2}-\frac{1}{(n-1)^2})=\sum\limits_{n=2}^{\infty}\frac{1}{n^2}-\sum\limits_{n=2}^{\infty}{\frac{1}{(n-1)^2}}$$
then you can see that $$\sum\limits_{n=2}^{\infty}\frac{1}{n^2}=\sum\limits_{n=1}^{\infty}\frac{1}{n^2}-(\frac{1}{(n=1)^2})=\sum\limits_{n=1}^{\infty}\frac{1}{n^2}-1$$
and at the same time that
$$\sum\limits_{n=2}^{\infty}{\frac{1}{(n-1)^2}}=\sum\limits_{n=1}^{\infty}\frac{1}{n^2}$$
then using what has already been said and this sum ,discovered by euler,$ \sum\limits_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}$ We can replace it and gives us
$$\sum\limits_{n=2}^{\infty}\frac{1}{n^2}-\sum\limits_{n=2}^{\infty}{\frac{1}{(n-1)^2}}=\frac{\pi^2}{6}-1-\frac{\pi^2}{6}=-1$$
2)Also in the same way we can generalize these sums for any exponent
$$\sum\limits_{n=2}^{\infty}(\frac{1}{n^k}-\frac{1}{(n-1)^k})$$
Applying the same steps as before we get
$$\sum\limits_{n=2}^{\infty}(\frac{1}{n^k}-\frac{1}{(n-1)^k})=\sum\limits_{n=2}^{\infty}\frac{1}{n^k}-\sum\limits_{n=2}^{\infty}{\frac{1}{(n-1)^k}}=\zeta(k)-1-\zeta(k)=-1$$
Thanks you for reading.