What is the sum from n=0 to infinity of 1/(4n+1) ?

The sum from n=0 to infinity of 1/(4n+1) is diverges

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sum from n=0 to infinity of 1/(4n+1)

n = 0 \sum \infty \frac{1}{4 n + 1}

diverges

Solution steps

n = 0 \sum \infty \frac{1}{4 n + 1}

Apply Series Integral Test:

diverges

= diverges

\int \frac{1}{e ^{x}} d x

y^{^{''}} + 4 y^{^{'}} + 2 y = 0

\frac{\partial}{\partial u} (tan (\frac{u}{v}))

\int \frac{x}{( 3 x ^{2} - 4 ) ^{2}} d x

4 x^{2}, 16

What is the sum from n=0 to infinity of 1/(4n+1) ?
The sum from n=0 to infinity of 1/(4n+1) is diverges