What is the sum from n=1 to infinity of (1+c)^{-n} ?

The sum from n=1 to infinity of (1+c)^{-n} is converges

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sum from n=1 to infinity of (1+c)^{-n}

n = 1 \sum \infty (1 + c)^{- n}

converges

Solution steps

n = 1 \sum \infty (1 + c)^{- n}

n \to \infty lim (n (\frac{( 1 + c ) ^{- n}}{( 1 + c ) ^{- (n + 1)}} - 1)) = \infty

L > 1,

by the Raabe's test

= converges

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

n = 1 \sum \infty \frac{n}{n ( n + 1 )}

n = 1 \sum \infty \frac{8}{n !}

n = 1 \sum \infty cos (\frac{19}{n})

n = 0 \sum \infty (sin (7))^{n}

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