What is the sum from n=1 to infinity of 2-1/n ?

The sum from n=1 to infinity of 2-1/n is diverges

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sum from n=1 to infinity of 2-1/n

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

diverges

Solution steps

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

Multiply by the conjugate of

2 - \frac{1}{n} : \frac{4 - \frac{1}{n ^{2}}}{2 + \frac{1}{n}}

= n = 1 \sum \infty \frac{4 - \frac{1}{n ^{2}}}{2 + \frac{1}{n}}

Simplify

\frac{4 - \frac{1}{n ^{2}}}{2 + \frac{1}{n}} : \frac{2 n - 1}{n}

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

Apply Series Divergence Test:

diverges

= diverges

n = 8 \sum \infty ln (\frac{n}{4})

k = 1 \sum \infty (cos (18))^{k}

n = 0 \sum \infty (sin (9))^{n}

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

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

What is the sum from n=1 to infinity of 2-1/n ?
The sum from n=1 to infinity of 2-1/n is diverges