What is the sum from n=1 to infinity of (-1)^n(3n)/(7n^2+1) ?

The sum from n=1 to infinity of (-1)^n(3n)/(7n^2+1) is converges

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sum from n=1 to infinity of (-1)^n(3n)/(7n^2+1)

n = 1 \sum \infty (- 1)^{n} \frac{3 n}{7 n ^{2} + 1}

converges

Solution steps

n = 1 \sum \infty (- 1)^{n} \frac{3 n}{7 n ^{2} + 1}

Apply the constant multiplication rule:

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

= 3 \cdot n = 1 \sum \infty (- 1)^{n} \frac{n}{7 n ^{2} + 1}

\frac{n}{7 n ^{2} + 1} = \frac{( n ) ( 7 n ^{2} - 1 )}{( 7 n ^{2} ) ^{2} - ( 1 ) ^{2}}

= 3 \cdot n = 1 \sum \infty (- 1)^{n} \frac{n ( 7 n ^{2} - 1 )}{( 7 n ^{2} ) ^{2} - 1 ^{2}}

Simplify

(- 1)^{n} \frac{n ( 7 n ^{2} - 1 )}{( 7 n ^{2} ) ^{2} - 1 ^{2}} : \frac{( - 1 ) ^{n} n}{7 n ^{2} + 1}

= 3 \cdot n = 1 \sum \infty \frac{( - 1 ) ^{n} n}{7 n ^{2} + 1}

Apply Alternating Series Test:

converges

= 3 converges

= converges

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

n = 5 \sum \infty \frac{28}{n + 2}

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

n = 1 \sum \infty \frac{9 ^{n} + n}{n !}

n = 0 \sum \infty e^{5 n}

What is the sum from n=1 to infinity of (-1)^n(3n)/(7n^2+1) ?
The sum from n=1 to infinity of (-1)^n(3n)/(7n^2+1) is converges