What is the sum from k=1 to infinity of (k-5)/(k+2) ?

The sum from k=1 to infinity of (k-5)/(k+2) is diverges

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sum from k=1 to infinity of (k-5)/(k+2)

k = 1 \sum \infty \frac{k - 5}{k + 2}

diverges

Solution steps

k = 1 \sum \infty \frac{k - 5}{k + 2}

Apply Series Divergence Test:

diverges

= diverges

n = 0 \sum \infty \frac{1}{n \cdot ( n + 1 )}

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

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

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

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

What is the sum from k=1 to infinity of (k-5)/(k+2) ?
The sum from k=1 to infinity of (k-5)/(k+2) is diverges