Convergence of $\sum_{n=1}^\infty\left(\frac{1}{a_n^2}-\frac{1}{a_{n+1}^2}\right)$ where $a_1=1$ and $a_n=2-\frac 1n$ for $n\geq 2$ Let $a_1=1\ $ and $\ a_n=2-\frac 1n$ for $n\geq 2$. Then $$\sum_{n=1}^\infty\bigg(\frac{1}{a_n^2}-\frac{1}{a_{n+1}^2}\bigg)$$
converges to?

I started by expanding the sum:
$\sum_{n=1}^\infty\bigg(\frac{1}{a_n^2}-\frac{1}{a_{n+1}^2}\bigg)=\bigg(\frac{1}{a_1^2}-\frac{1}{a_2^2}\bigg)+\bigg(\frac{1}{a_2^2}-\frac{1}{a_3^2}\bigg)+\bigg(\frac{1}{a_3^2}-\frac{1}{a_4^2}\bigg)+\ldots$
$=\frac{1}{a_1^2}+\bigg(-\frac{1}{a_2^2}+\frac{1}{a_2^2}\bigg)+\bigg(-\frac{1}{a_3^2}+\frac{1}{a_3^2}\bigg)+\bigg(-\frac{1}{a_4^2}+\frac{1}{a_4^2}\bigg)+\ldots$
$=\frac{1}{a_1^2}$
$=1$

Now, the strange part is that the answer to this question is known to me as $0.75$. Is there something wrong with what I did? How does one arrive at $0.75$ as a solution? 
PS: I am aware that infinite sums can behave in strange ways, and sometimes more than one solution may follow logically. Is this one such case?

Let $$S_N=\sum_{n=1}^N\left(\frac{1}{a_n^2}-\frac{1}{a_{n+1}^2}\right)=\frac{1}{a_1^2}-\frac{1}{a_{N+1}^2}.$$
Since $a_1=1$ and $a_n\to 2$, we have 
$$\lim_{N\to\infty}S_N=1-\frac{1}{2^2}=\frac{3}{4}.$$
