What is the sum from n=1 to infinity of 8-5/(n^2) ?

The sum from n=1 to infinity of 8-5/(n^2) is diverges

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sum from n=1 to infinity of 8-5/(n^2)

n = 1 \sum \infty 8 - \frac{5}{n ^{2}}

diverges

Solution steps

n = 1 \sum \infty 8 - \frac{5}{n ^{2}}

Apply Series Divergence Test:

diverges

= diverges

n = 5 \sum \infty ln (1 + \frac{1}{n})

n = 1 \sum \infty \frac{5}{n ln ( 3 n )}

n = 1 \sum \infty \frac{9}{n !}

n = 0 \sum \infty \frac{n}{5 ^{n}}

n = 1 \sum \infty \frac{1}{n + 8 ^{n}}

What is the sum from n=1 to infinity of 8-5/(n^2) ?
The sum from n=1 to infinity of 8-5/(n^2) is diverges