What is the sum from n=1 to infinity of 1-1/(n^2) ?

The sum from n=1 to infinity of 1-1/(n^2) is diverges

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sum from n=1 to infinity of 1-1/(n^2)

n = 1 \sum \infty 1 - \frac{1}{n ^{2}}

diverges

Solution steps

n = 1 \sum \infty 1 - \frac{1}{n ^{2}}

Apply Series Divergence Test:

diverges

= diverges

n = 0 \sum \infty \frac{3 ^{n}}{7 ^{n}}

n = 0 \sum \infty \frac{2 n - 1}{2 ^{n}}

n = 1 \sum \infty \frac{10}{n ^{2}}

n = 1 \sum \infty \frac{1}{ln ( n + 2 )}

n = 1 \sum \infty \frac{n}{( n + 1 ) ^{2}}

What is the sum from n=1 to infinity of 1-1/(n^2) ?
The sum from n=1 to infinity of 1-1/(n^2) is diverges