What is the sum from k=1 to infinity of 5/((k+4)^2) ?

The sum from k=1 to infinity of 5/((k+4)^2) is converges

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sum from k=1 to infinity of 5/((k+4)^2)

k = 1 \sum \infty \frac{5}{( k + 4 ) ^{2}}

converges

Solution steps

k = 1 \sum \infty \frac{5}{( k + 4 ) ^{2}}

Apply the constant multiplication rule:

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

= 5 \cdot k = 1 \sum \infty \frac{1}{( k + 4 ) ^{2}}

Apply Series Comparison Test:

converges

= 5 converges

= converges

n = 0 \sum \infty n^{\frac{1}{3}}

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

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

n = 0 \sum \infty \frac{3}{n ^{2} + 3}

n = 0 \sum \infty \frac{3}{n ^{2} + 4}

What is the sum from k=1 to infinity of 5/((k+4)^2) ?
The sum from k=1 to infinity of 5/((k+4)^2) is converges