What is the sum from k=1 to infinity} 1/(sqrt(k) of-1/(\sqrt{k+2)) ?

The sum from k=1 to infinity} 1/(sqrt(k) of-1/(\sqrt{k+2)) is converges

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sum from k=1 to infinity} 1/(sqrt(k) of-1/(\sqrt{k+2))

k = 1 \sum \infty \frac{1}{k} - \frac{1}{k + 2}

converges

Solution steps

k = 1 \sum \infty \frac{1}{k} - \frac{1}{k + 2}

Apply Series Integral Test:

converges

= converges

n = 0 \sum \infty e^{- 19 n}

n = 7 \sum \infty \frac{n !}{n - 6}

k = 1 \sum \infty \frac{1}{k ^{2} + 3 k + 2}

n = 2 \sum \infty \frac{10}{ln ( n )}

n = 0 \sum \infty \frac{9}{2} (\frac{1}{2})^{n}

What is the sum from k=1 to infinity} 1/(sqrt(k) of-1/(\sqrt{k+2)) ?
The sum from k=1 to infinity} 1/(sqrt(k) of-1/(\sqrt{k+2)) is converges