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

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

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

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

diverges

Solution steps

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

Apply Series Limit Comparison Test:

diverges

= diverges

n = 0 \sum \infty \frac{n}{9 + n ^{2}}

n = 0 \sum \infty 7 (\frac{2}{3})^{n + 2}

n = 1 \sum \infty \frac{e ^{\frac{8}{n}}}{n}

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

n = 0 \sum \infty e^{- 7 n}

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