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

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

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

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

diverges

Solution steps

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

Apply Series Divergence Test:

diverges

= diverges

n = 0 \sum \infty (\frac{3}{π})^{n}

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

n = 0 \sum \infty 8 \cdot (\frac{5}{8})^{n}

n = 0 \sum \infty 5 + \frac{15}{5} + \frac{45}{16}

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

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