What is the sum from n=1 to infinity of (10)/(n^2) ?

The sum from n=1 to infinity of (10)/(n^2) is converges

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sum from n=1 to infinity of (10)/(n^2)

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

converges

Solution steps

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

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

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

Apply p-Series Test:

converges

= 10 converges

= converges

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

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

n = 1 \sum \infty (\frac{2}{n})^{4}

n = 1 \sum \infty \frac{3}{n ^{2} + 3}

n = 0 \sum \infty n^{^{'}} + n

What is the sum from n=1 to infinity of (10)/(n^2) ?
The sum from n=1 to infinity of (10)/(n^2) is converges