fabianmartinOTQ

2022-12-04

Consider applying the Limit Comparison Test on $S=\sum _{k=1}^{\mathrm{\infty }}\left(\frac{3k}{5\left(k-1\right)\left(k+1\right)}\right)$
a) Setup and calculate the limit in this test, comparing this series to $\sum _{k=2}^{\mathrm{\infty }}\left(\frac{1}{k}\right)$
b) Interpret the result of this limit in this test.

hayaniqWf

Expert

$S=\sum _{k=L}^{\mathrm{\infty }}\frac{3k}{5\left(k-1\right)\left(k+1\right)}=\sum {a}_{k}\phantom{\rule{0ex}{0ex}}\sum {b}_{R}=\sum \frac{1}{R}$
a)$l{t}_{k\to \mathrm{\infty }}\frac{{a}_{R}}{{b}_{R}}=l{t}_{k\to \mathrm{\infty }}\frac{3{k}^{2}}{5\left(k-1\right)\left(k+1\right)}=l{t}_{k\to \mathrm{\infty }}\frac{3}{5|1-\frac{1}{R}||1+\frac{1}{R}|}=\frac{3}{5}$
b) By limit compasion test, since $\sum {b}_{R}$ is divergent
$\sum _{k=2}^{\mathrm{\infty }}\frac{3k}{5\left(k-1\right)\left(k+1\right)}$ is divergent

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