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

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

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

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

diverges

Solution steps

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

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

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

Apply Series Divergence Test:

diverges

= diverges

x e^{- 0.2 x}, y = 0, x = 1

x \to 0 lim (e^{π (2 + 1)})

\int \frac{1}{t ^{3} t ^{2} - 9} d t

\int tan^{5} (2 x) sec^{2} (2 x) d x

\frac{4}{1 - 2 x ^{3}}

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