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

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

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

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

diverges

Solution steps

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

Apply Series Divergence Test:

diverges

= diverges

2 y^{^{''}} + 5 y^{^{'}} - 3 y = 0, y (0) = 1, y^{^{'}} (0) = 4

n = 0 \sum \infty \frac{1}{3 ^{n + 1}}

\int tan^{3} (2 t) sec^{3} (2 t) d t

\frac{d ^{2}}{d x ^{2}} (cos^{2} (2 π x))

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