What is the sum from n=5 to infinity of e^{3-4n} ?

The sum from n=5 to infinity of e^{3-4n} is converges

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sum from n=5 to infinity of e^{3-4n}

n = 5 \sum \infty e^{3 - 4 n}

converges

Solution steps

n = 5 \sum \infty e^{3 - 4 n}

Apply Series Ratio Test:

converges

= converges

y^{^{''}} + 4 π^{2} y = cos^{3} (2 π x)

\int 2 x I n x^{2} d x

\frac{d}{d x} (\frac{8 ^{x}}{7 ^{x}})

\int \frac{v}{v + 2} d v

\int \frac{x ^{2}}{9 + 4 x ^{2}} d x

What is the sum from n=5 to infinity of e^{3-4n} ?
The sum from n=5 to infinity of e^{3-4n} is converges