What is the sum from n=0 to infinity of 1/(5^n-4^n) ?

The sum from n=0 to infinity of 1/(5^n-4^n) is converges

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sum from n=0 to infinity of 1/(5^n-4^n)

n = 0 \sum \infty \frac{1}{5 ^{n} - 4 ^{n}}

converges

Solution steps

n = 1 \sum \infty \frac{1}{5 ^{n} - 4 ^{n}}

Factor

5^{n} - 4^{n} : - (- 5^{n} + 4^{n})

= n = 1 \sum \infty \frac{1}{- ( - 5 ^{n} + 4 ^{n} )}

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= - n = 1 \sum \infty \frac{1}{- 5 ^{n} + 4 ^{n}}

Apply Series Ratio Test:

converges

= - converges

= converges

n = 1 \sum \infty (cos (10))^{n}

n = 1 \sum \infty 7 n^{2} e^{- n^{3}}

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

n = 1 \sum \infty \frac{1}{12 n + 11}

n = 3 \sum \infty (\frac{1}{4})^{n}

What is the sum from n=0 to infinity of 1/(5^n-4^n) ?
The sum from n=0 to infinity of 1/(5^n-4^n) is converges