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

The sum from n=1 to infinity of (4n)/(n^2+1) is diverges

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

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

diverges

Solution steps

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

Apply the constant multiplication rule:

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

= 4 \cdot n = 1 \sum \infty \frac{n}{n ^{2} + 1}

Apply Series Limit Comparison Test:

diverges

= 4 diverges

= diverges

n = 0 \sum \infty 8 (\frac{2}{3})^{n}

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

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

n = 1 \sum \infty \frac{7}{n !}

n = 0 \sum \infty (\frac{7}{6})^{n}

What is the sum from n=1 to infinity of (4n)/(n^2+1) ?
The sum from n=1 to infinity of (4n)/(n^2+1) is diverges