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

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

Popular >

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

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

diverges

Solution steps

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

Factor

1 - n : - (- 1 + n)

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

Apply the constant multiplication rule:

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

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

Expand

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

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

Apply Series Limit Comparison Test:

diverges

= - diverges

= diverges

n = 0 \sum \infty - ln (3 - n)

n = 1 \sum \infty (0.6)^{n}

k = 0 \sum \infty \frac{k}{8 k + 1}

n = 2 \sum \infty [\frac{2}{5 ^{n}}]

n = 1 \sum \infty 3 (- \frac{1}{5})^{3 n}

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