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

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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}}

Apply Series Comparison Test:

diverges

= diverges

n = 6 \sum \infty \frac{1}{( n - 3 ) ^{4}}

n = 1 \sum \infty \frac{1}{n ^{7} + 8}

n = 0 \sum \infty \frac{3}{- 3 ^{n}}

n = 1 \sum \infty \frac{5 ^{n}}{n + 1}

k = 1 \sum \infty \frac{6}{k}

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