What is the sum from n=0 to infinity of e^{2ln(n)} ?

The sum from n=0 to infinity of e^{2ln(n)} is diverges

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sum from n=0 to infinity of e^{2ln(n)}

n = 0 \sum \infty e^{2 l n (n)}

diverges

Solution steps

n = 1 \sum \infty e^{2 l n (n)}

Simplify

e^{2 l n (n)} : n^{2}

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

Apply Series Divergence Test:

diverges

= diverges

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

n = 0 \sum \infty 5 (\frac{3}{π})^{n}

n = 1 \sum \infty (\frac{- 1}{7})^{n}

n = 0 \sum \infty (\frac{1}{2})^{n - 2}

n = 0 \sum \infty (\frac{- 1}{4})^{n}

What is the sum from n=0 to infinity of e^{2ln(n)} ?
The sum from n=0 to infinity of e^{2ln(n)} is diverges