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

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

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sum from n=1 to infinity of 1/(n^2)-1/n

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

diverges

Solution steps

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

Factor

\frac{1}{n ^{2}} - \frac{1}{n} : - (- \frac{1}{n ^{2}} + \frac{1}{n})

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

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

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

Apply Series Limit Comparison Test:

diverges

= - diverges

= diverges

x \to 5 lim (x^{2} - 25)

\int \frac{2}{( x ^{2} - 3 x + 2 )} d x

d er i v a t i v e f (t) = \frac{t}{t ^{2} + 7}

\frac{d}{d x} (e^{x} (6 x^{2} - 12 x + 12))

\frac{d}{d y} (\frac{1}{y ^{2}})

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