What is the sum from n=9 to infinity of e^{3-4n} ?

The sum from n=9 to infinity of e^{3-4n} is converges

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sum from n=9 to infinity of e^{3-4n}

n = 9 \sum \infty e^{3 - 4 n}

converges

Solution steps

n = 9 \sum \infty e^{3 - 4 n}

Apply Series Ratio Test:

converges

= converges

y = cos^{4} (x) - sin^{4} (x)

\int sin^{3} (x) cos (d) x d x

x^{2} - 2 x + 4

\int (7 x^{3} + 4 x^{2}) d x

f (x) = 4 x - x^{2}

What is the sum from n=9 to infinity of e^{3-4n} ?
The sum from n=9 to infinity of e^{3-4n} is converges