What is the sum from n=3 to infinity of (1+n)^{-1} ?

The sum from n=3 to infinity of (1+n)^{-1} is diverges

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sum from n=3 to infinity of (1+n)^{-1}

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

diverges

Solution steps

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

Apply exponent rule:

a^{- 1} = \frac{1}{a}

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

Apply Series Limit Comparison Test:

diverges

= diverges

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

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

n = 0 \sum \infty \frac{1}{( n + 2 ) n !}

n = 1 \sum \infty \frac{2 n + 1}{6 ^{n}}

n = 0 \sum \infty \frac{4}{3 - n}

What is the sum from n=3 to infinity of (1+n)^{-1} ?
The sum from n=3 to infinity of (1+n)^{-1} is diverges