What is the sum from n=1 to infinity of 7/(n^2+25) ?

The sum from n=1 to infinity of 7/(n^2+25) is converges

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sum from n=1 to infinity of 7/(n^2+25)

n = 1 \sum \infty \frac{7}{n ^{2} + 25}

converges

Solution steps

n = 1 \sum \infty \frac{7}{n ^{2} + 25}

Apply the constant multiplication rule:

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

= 7 \cdot n = 1 \sum \infty \frac{1}{n ^{2} + 25}

Apply Series Comparison Test:

converges

= 7 converges

= converges

\int_{0}^{\frac{1}{2}} \frac{5}{1 + 4 x ^{2}} d x

e^{x} (3 x^{2} - x + 1)

\frac{\partial}{\partial y} (e^{- x y^{2}} + y^{3} x^{4})

x y

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

What is the sum from n=1 to infinity of 7/(n^2+25) ?
The sum from n=1 to infinity of 7/(n^2+25) is converges