What is the sum from n=1 to infinity of 1/(n+9^n) ?

The sum from n=1 to infinity of 1/(n+9^n) is converges

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sum from n=1 to infinity of 1/(n+9^n)

n = 1 \sum \infty \frac{1}{n + 9 ^{n}}

converges

Solution steps

n = 1 \sum \infty \frac{1}{n + 9 ^{n}}

Apply Series Ratio Test:

converges

= converges

n = 0 \sum \infty \frac{9}{10}

n = 2 \sum \infty 4 n^{- 2}

n = 0 \sum \infty (\frac{1}{13})^{n}

n = 6 \sum \infty \frac{1}{( n - 5 ) ^{2}}

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

What is the sum from n=1 to infinity of 1/(n+9^n) ?
The sum from n=1 to infinity of 1/(n+9^n) is converges