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

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

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sum from n=1 to infinity of (((3n-1))/((3n)))^{n^2}

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

converges

Solution steps

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

Apply Series Root Test:

converges

= converges

tan (x) sin^{2} (y) d x + cos^{2} (x) cot (y) d y = 0

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

y^{^{'}} = 32 x^{3} y^{3}

\int_{- 2}^{2} x^{2} e^{x^{3} + 8} d x

5 ln (x)

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