Does $\sum{\ln n\over n^{4/3}}$ converge or diverge?

By the integral test you may compare your series with the following integral: $$ \int_1^\infty \frac{\ln x}{x^{4/3}} dx=\left.\frac{x^{-4/3+1}}{-4/3+1}\ln x\right|_1^\infty-\frac1{-4/3+1}\int_1^\infty x^{-4/3+1}\frac1{x} dx=9<\infty $$ giving the convergence of your series $$ \sum_1^\infty\frac{\ln n}{n^{4/3}}. $$


Let's apply Cauchy condensation test :

$$2^n a_{2^n}=2^n\cdot \frac{\log 2^n}{2^{4n/3}}=\log{2}\cdot \frac{n}{2^{n/3}}$$

$\sum_n^\infty \frac{n}{2^{n/3}} $ clearly converges, so the given series also converges.

Tags:

Integration

Calculus

Sequences And Series

Divergent Series

Convergence Divergence

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