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

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

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

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

converges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Comparison Test:

converges

= 2 converges

= converges

n = 0 \sum \infty (- 1)^{n} 5^{n}

n = 1 \sum \infty \frac{1}{6 n + 1}

n = 0 \sum \infty (\frac{8 π}{e})^{n}

n = 1 \sum \infty (cos (20))^{n}

x = 1 \sum \infty \frac{x}{e ^{x}}

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