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

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

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

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

converges

Solution steps

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

Simplify

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

Apply p-Series Test:

converges

= converges

k = 1 \sum \infty \frac{1}{k ^{2} + 5 k + 6}

n = 1 \sum \infty 3 (0.1)^{n - 1}

n = 1 \sum \infty 6 (0.4)^{n - 1}

n = 2 \sum \infty \frac{9}{n ln ( n )}

n = 0 \sum \infty (- \frac{7}{9})^{n}

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